Answer: 

(i) Mass of one electron = 9.10939 × 10^–31 kg
Number of electrons that weigh 9.10939 × 10^–31 kg = 1
Number of electrons that will weigh 1 g (1 × 10^–3 kg)

= \(\frac{1}{9.10939 × 10^{-31}kg}\) × 1 × 10^–3 kg

= = 0.1098 × 10^{–3 + 31}
= 0.1098 × 10²⁸
= 1.098 × 10²⁷
(ii) Mass of one electron = 9.10939 × 10^–31 kg
Mass of one mole of electron = (6.022 × 10²³) × (9.10939 ×10^–31 kg)
= 5.48 × 10^–7 kg
Charge on one electron = 1.6022 × 10^–19 coulomb
Charge on one mole of electron = (1.6022 × 10^–19 C) (6.022 × 10²³)
= 9.65 × 10⁴ C

Answer: 

Answer 2.2:
(i) Number of electrons present in 1 molecule of methane (CH₄) = 1 × (6) + 4 × (1) = 10
Number of electrons present in 1 mole i.e., 6.023 × 10²³ molecules of methane.

= 6.022 × 10²³ × 10 = 6.022 × 1024
(ii) (a) Number of atoms of 14C in 1 mole= 6.023 × 10²³
Since 1 atom of ¹⁴C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of ¹⁴C is (6.023 × 10²³) ×8. Or, 14 g of ¹⁴C contains (6.022 × 10²³ × 8) neutrons.
Number of neutrons in 7 mg

= \(\frac{6.022 × 10{23} × 8 × 7 mg }{14000 mg}\)

= 2.4092 × 10²¹

(b) Mass of one neutron = 1.67493 × 10^–27 kg
Mass of total neutrons in 7 g of ¹⁴C
= (2.4092 × 10²¹) (1.67493 × 10^–27 kg)
= 4.0352 × 10^–6 kg

(iii) (a) 1 mole of NH₃ = 1 × (14) + 3 × (1)} g of NH₃
= 17 g of NH3
= 6.022 × 10²³ molecules of NH₃
Total number of protons present in 1 molecule of NH₃
= 1×(7) + 3×(1)
= 10

Number of protons in 6.023 × 10²³ molecules of NH₃
= (6.023 × 10²³) ×  (10)
= 6.023 × 10²⁴
⇒ 17 g of NH₃ contains (6.023 × 10²⁴) protons. Number of protons in 34 mg of NH₃

= \(\frac{6.022 × 10{24} × 34 mg }{17000 mg}\)

= 1.2046 × 10²²

(b) Mass of one proton = 1.67493 × 10^–27 kg
Total mass of protons in 34 mg of NH₃
= (1.67493 × 10^–27 kg) (1.2046 × 10²²)
= 2.0176 × 10^–5 kg

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Answer: 

\(_{6}^{13}C\)

Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7

\(_{8}^{16}O\)

Atomic mass = 16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8

\(_{12}^{24}Mg\)

Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24 – 12 = 12

\(_{56}^{26}Fe\)

Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30

\(_{88}^{38}Sr\)

Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50

Answer: 

(i) \(_{17}^{35}Cl\)   (ii) \(_{92}^{233}U\)  (iii) \(_{4}^{9}Be\)

Answer: 

From the expression

λ = c/v 

We get, v = c/ λ ……….. (i)

Where, ν = frequency of yellow light 

velocity of light in vacuum = 3 × 10⁸ m/s
λ = wavelength of yellow light = 580 nm = 580 × 10^–9 m

Substituting the values in expression (i):

v = ( 3 × 10⁸ m/s) / (580 × 10^–9 m) = 5.17 × 10¹⁴ s ^-1

Thus, frequency of yellow light emitted from the sodium lamp = 5.17 × 10¹⁴ s ^-1
Wave number of yellow light

\(\bar v\) = 1 / λ = 1 / (580 × 10^–9 m) = 1.72 × 10⁶ m^-1

Answer: 

Answer 2.6:
(i) Energy (E) of a photon is given by the expression, E = hv

Where, h = Planck’s constant = 6.626 × 10^–34 Js
ν = frequency of light = 3 × 10¹⁵ Hz
Substituting the values in the given expression of E:
E = (6.626 × 10^–34) (3 × 10¹⁵)
E = 1.988 × 10^–18 J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression

E= hc/ λ

h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s
Substituting the values in the given expression of E:

E = (6.626 × 10^–34 Js) (3 × 10⁸ m/s)/(0.50 × 10^–15)

= 3.976 × 10^–15 J

∴ E = 3.98 × 10^–15 J

Answer: 

Frequency (ν) of light = \(\frac{1}{period}\)

= \(\frac{1}{2.0 × 10^{-10}s}\)

= 5 × 10⁹ s^-1

Wavelength (λ ) of light =  \(\frac{c}{v}\)

Where,
c = velocity of light in vacuum = 3×10⁸ m/s
Substituting the value in the given expression of λ:

λ = \(\frac{3 × 10^8}{5.0 × 10^{9}s}\) = 6×10^-2 m

Wave number \(\bar v\) of light = 1/ λ = 1/ 6×10^-2 m

= 1.66×10¹ m^-1   

= 16.66 m^-1 

Answer: 

Energy (E) of a photon = hν
Energy (E_n) of ‘n’ photons = nhν

n= \(\frac{E_nλ}{hc}\)

Where, λ = wavelength of light = 4000 pm = 4000 ×10^–12 m

c = velocity of light in vacuum = 3 × 10⁸ m/s

h = Planck’s constant = 6.626 × 10^–34 Js

Substituting the values in the given expression of n:

n= \(\frac{1 × 4000 × 10^{-12} }{6.626 × 10^{-34} × 3 × 10^8 }\)

= 2.012 × 10¹⁶

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are
2.012 × 10¹⁶

Answer: 

(i) Energy (E) of a photon = hν = hc/ λ

Where, h = Planck’s constant = 6.626 × 10^–34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of photon = 4 × 10^–7 m
Substituting the values in the given expression of E:

E = \(\frac{6.626 × 10^{-34} × 3 × 10^8 }{4 × 10^{-7} }\)

= 4.9695 × 10^-19 J

Hence, the energy of the photon is 4.97 × 10^–19 J.

(ii) The kinetic energy of emission Ek is given by

= hv -hv₀

= (E-W)eV

= ( \(\frac{4.6965 × 10^{-19} }{1.6020 × 10^{-19} }\) – 2.13 ) eV

= (3.1020 – 2.13) eV
= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (ν) can be calculated by the expression.

\(\frac{1}{2} mv^2\) = hv -hv₀

v = \(\sqrt{\frac{2(hv-hv_0)}{m}}\)

where ( hv-hv₀ ) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron

(ref – DISCOVERY OF SUB-ATOMIC PARTICLES, ELECTRON, PROTONS AND NEUTRONS for mass of electron).

Substituting the values in the given expression of v:

v = \(\sqrt{\frac{0.9720 × 1.6020 × 10^{-19}J }{9.01939 × 10^{-31}kg}}\)

v = \(\sqrt{0.3418 × 10^{12}m^2s^{-2}}\)

v = 5.84 × 10⁵ ms^–1

Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^–1.

Answer: 

Energy of sodium (E) = \(\frac{N_Ahc}{λ}\)

= \(\frac{(6.023 × 10^{23}mol^{-1})(6.626 × 10^{-34}Js)(3  × 10^8ms^{-1}) }{242  × 10^{-9}m}\)

= 4.947 × 10⁵ J mol^–1
= 494.7 × 10³ J mol^–1
= 494 kJ mol^–1

Answer: 

Power of bulb, P = 25 Watt = 25 Js^–1
Energy of one photon, E = hν = \(\frac{hc}{λ}\)

= Substituting the values in the given expression of E:

= \(\frac{(6.626 × 10^{-34}Js)(3  × 10^8ms^{-1}) }{0.57  × 10^{-6}m}\)

E = 34.87 × 10^–20 J
Rate of emission of quanta per second

= P/E

= \(\frac{25 }{34.87  × 10^{-20}}\)

= 7.169 x 10¹⁹s^-1

Answer: 

Threshold wavelength of radiation λ_{0 = 6800 Å  =  6800 × 10^–10 m}

Threshold frequency v₀ of the metal = c /λ₀

_{= \(\frac{3  × 10^8ms^{-1}}{6800  × 10^{-10}m}\)}

= 4.41 × 10¹⁴ s^–1
Thus, the threshold frequency v₀ of the metal is 4.41 × 10¹⁴ s^–1

Hence, work function (W₀) of the metal = hν₀
= (6.626 × 10^–34 Js) (4.41 × 1014 s^–1)
= 2.922 × 10^–19 J

Answer: 

The n_i = 4 to n_f = 2 transition will give rise to a spectral line of the Balmer series.

The energy involved in the transition is given by the relation,

E = 2.18 × 10^–18 ×[\(\frac{1}{n_i^2} – \frac{1}{n_f^2}\)]

Substituting the values in the given expression of E:

= 2.18 × 10^–18 ×[\(\frac{1}{4^2} – \frac{1}{2^2}\)]

= 2.18 × 10^–18 ×[\(-\frac{3}{16}\)]

E = – (4.0875 × 10^–19 J)
The negative sign indicates the energy of emission.
Wavelength of light emitted

λ =  \(\frac{hc}{E}\)  …….( since E = hν = \(\frac{hc}{λ}\))

Substituting the values in the given expression of λ: 

= \(\frac{(6.626 × 10^{-34}Js)(3  × 10^8ms^{-1}) }{4.0875  × 10^{-19}J}\)

= 4.8631 x 10 ^-7 m

= 486.31 x 10 ^-9 m

= 486 nm

Answer: 

The expression of energy is given by,

\(\frac{(-2.18 × 10^{-18}Z^2) }{n^2}\)

Where,
Z = atomic number of the atom n = principal quantum number
For ionization from n₁ = 5 to n₂ = ∞ 

ΔE = E_∞ -E₅

= \(\frac{(-2.18 × 10^{-18}J )× (1^2) }{∞^2}\) – \(\frac{(-2.18 × 10^{-18}J )× (1^2) }{5^2}\)

=  \(\frac{(-2.18 × 10^{-18}J )× (1^2) }{5^2}\) ……… ( since 1/∞ = 0)

= 0.0872 x 10 ^-18 J

ΔE = 0.0872 x 10 ^-18 J

Hence, the energy required for ionization from n = 5 to n =  ∞ is 8.72 x 10 ^-20 J

Energy required for n₁ = 1 to n =  ∞  

ΔE = E_∞ -E₁

= \(\frac{(-2.18 × 10^{-18}J )× (1^2) }{∞^2}\) – \(\frac{(-2.18 × 10^{-18}J )× (1^2) }{1^2}\)

= 2.18 x 10 ^-18 J

Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

Answer: 

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by 

= n(n-1)/2

Given, n = 6

Number of spectral lines = 6(6-1)/2 = 15

Answer: 

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

E₅ =  –2.18 × 10^–18 J/(5)²

E₅ = –8.72 × 10^–20 J

(ii) Radius of Bohr’s n^th orbit for hydrogen atom is given by, r_n = (0.0529 nm) n²
For, n = 5
r₅ = (0.0529 nm)  × (5)²
r₅ = 1.3225 nm

Answer: 

For the Balmer series, n_i = 2. Thus, the expression of wave number \(\bar v\) is given by,

\(\bar v\) = \(\frac{1}{2^2} – \frac{1}{n_f^2}\) × 1.097 x 10⁷ m^-1

wave number \(\bar v\) is inversely proportional to wavelength of transition.

Hence, for the longest wavelength transition \(\bar v\)  has to be the smallest.

For \(\bar v\) to be minimum, n_f should be minimum.

For the Balmer series, a transition from n_i = 2 to n_f = 3 is allowed.
Hence, taking n_f = 3, we get:

\(\bar v\) = \(\frac{1}{2^2} – \frac{1}{3^2}\) × 1.097 x 10⁷ m^-1

= \(\frac{9-4}{36} \) × 1.097 x 10⁷ m^-1

\(\bar v\) = 1.5236 ×10⁶ m^-1

Answer: 

Energy (E) of the nth Bohr orbit of an atom is given by,

E_n = \(\frac{(-2.18 × 10^{-18}Z^2) }{n^2}\)

Where,
Z = atomic number of the atom
Ground state energy = – 2.18 × 10^–11 ergs
= –2.18 × 10^–11 × 10^–7 J
= – 2.18 × 10^–18 J
Energy required to shift the electron from n = 1 to n = 5 is given as:

ΔE = E₅ – E₁

= \(\frac{(-2.18 × 10^{-18} ×1^2) }{5^2}\) – (– 2.18 × 10^–18)

=  ( 2.18 × 10^–18) (\(\frac{(24 }{25}\))

= 2.0928 × 10^–18 J

wavelength of emitted light λ = \(\frac{hc}{E}\)

= \(\frac{6.626 × 10^{-34} × 3 × 10^8 }{2.0928 × 10^{-18} }\)

= 9.498 × 10^–8 m

Answer: 

Given E_n = (–2.18 × 10^–18)/n² J

Energy required for ionization from n = 2 is given by,

ΔE = E_∞ -E₂

= \((\frac{-2.18 × 10^{-18}}{∞^2}) -(\frac{-2.18 × 10^{-18}}{2^2}) \)J
= 0.545 × 10^–18 J

ΔE = 5.45 × 10^–19 J

 λ = hc/ΔE

Here, λ is the longest wavelength causing the transition.

=\( \frac{6.626 × 10^{-34} × 3 × 10^8 }{5.45 × 10^{-19} }\)

= 3647 × 10^–10 m
= 3647 Å

Answer: 

According to de Broglie’s equation, λ = \(\frac{h}{mv}\)

Where, λ = wavelength of moving particle m = mass of particle v = velocity of particle h = Planck’s constant
Substituting the values in the expression of λ:

λ = \(\frac{6.626 × 10^{-34}Js}{9.10939× 10^{-31}kg × 2.05 × 10^7ms^{-1}}\)

λ = 3.548 × 10^–11 m

Hence, the wavelength of the electron moving with a velocity of 2.05 × 10⁷ ms^–1  is 3.548 × 10^–11 m

Answer:

From de Broglie’s equation,

λ = h/mv

Given,
Kinetic energy (K.E) of the electron = 3.0 × 10^–25 J

Since K.E. = \(\frac{1}{2}mv^2\)

∴ Velocity v = \(\sqrt{2KE}{m}\)

= \(\sqrt{\frac{2× (3.0 × 10^{-25}J)}{9.01939 × 10^{-31}kg}}\) 

= \(\sqrt{6.5866 × 10^4}\)

v = 811.579 ms^-1

Substituting the value in the expression of λ:

= \(\frac{6.626 × 10^{-34} Js}{9.10939 × 10^{-31}kg  × 811.579 ms^{-1}   }\)

λ = 8.9625 × 10^–7 m

Hence, the wavelength of the electron is 8.9625 × 10^–7 m.

Answer:

Isoelectronic species have the same number of electrons.
Number of electrons in sodium (Na) = 11
Number of electrons in (Na⁺) = 10
A positive charge denotes the loss of an electron.
Similarly,
Number of electrons in K⁺ = 18
Number of electrons in Mg²⁺ = 10
Number of electrons in Ca²⁺ = 18
A negative charge denotes the gain of an electron by a species.
Number of electrons in sulphur (S) = 16
∴ Number of electrons in S^2- = 18
Number of electrons in argon (Ar) = 18
Hence, the following are isoelectronic species:
(1) Na⁺ and Mg²⁺ (10 electrons each)
(2) K⁺, Ca²⁺, S^2- and Ar (18 electrons each).

Answer:

(i) (a) H^– ion
The electronic configuration of H atom is 1s¹.
A negative charge on the species indicates the gain of an electron by it.
Electronic configuration of H^– = 1s²

(b) Na⁺ ion
The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.
A positive charge on the species indicates the loss of an electron by it.
Electronic configuration of Na⁺ = 1s² 2s² 2p⁶ 3s⁰ or 1s² 2s² 2p⁶

(c) O2^– ion
The electronic configuration of 0 atom is 1s² 2s² 2p⁴.
A dinegative charge on the species indicates that two electrons are gained by it.
Electronic configuration of O2^– ion = 1s² 2s² p⁶

(d) F^– ion
The electronic configuration of F atom is 1s² 2s² 2p⁵.
A negative charge on the species indicates the gain of an electron by it.
Electron configuration of F^– ion = 1s² 2s² 2p⁶

(ii) (a) 3s¹
Completing the electron configuration of the element as
1s² 2s² 2p⁶ 3s¹.
∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴ Atomic number of the element = 11

(c) 3p⁵
Completing the electron configuration of the element as
1s² 2s² 2p⁵.
∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
∴ Atomic number of the element = 9

(iii) (a) [He] 2s¹
The electronic configuration of the element is [He] 2s¹ = 1s² 2s¹.
∴ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(b) [Ne] 3s² 3p³
The electronic configuration of the element is [Ne] 3s² 3p³= 1s² 2s² 2p⁶ 3s² 3p³.
∴ Atomic number of the element = 15
∴ Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(c) [Ar] 4s² 3d¹
The electronic configuration of the element is [Ar] 4s² 3d¹= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹. Atomic number of the element = 21
∴ Hence, the element with the electronic configuration [Ar] 4s² 3d¹ is scandium (Sc).

Answer:

For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴ For l = 4, minimum value of n = 5

Answer:

For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2

Answer:

(i) For an atom to be neutral, the number of protons is equal to the number of electrons.
∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is
1s² 2s² 2p⁶ 3s² 3p6 4s² 3d¹⁰.

Answer:

H₂⁺

Number of electrons present in hydrogen molecule (H₂) = 1 + 1 = 2
∴ Number of electrons in H₂⁺ = 2 – 1 = 1
H₂:
Number of electrons in H₂ = 1 + 1 = 2

O₂⁺

Number of electrons present in oxygen molecule (O₂) = 8 + 8 = 16
∴ Number of electrons in O₂⁺ = 16 – 1 = 15

Answer:

(i) n = 3 (Given)
For a given value of n, l can have values from 0 to (n – 1).
∴ For n = 3
l = 0, 1, 2
For a given value of l, ml can have (2l + 1) values. For l = 0, m = 0 l = 1, m = – 1, 0, 1 l = 2, m = – 2, – 1, 0, 1, 2
For n = 3
l = 0, 1, 2 m₀ = 0
m₁ = – 1, 0, 1 m₂ = – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l = 2.
For a given value of l, ml can have (2l + 1) values i.e., 5 values.
∴ For l = 2 m₂ = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.
For p-orbital, l = 1.
For a given value of n, l can have values from zero to (n – 1).
For l is equal to 1, the minimum value of n is 2.
Similarly,
∴ For f-orbital, l = 4.
For l = 4, the minimum value of n is 5.
Hence, 1p and 3f do not exist.

Answer:

(a) n = 1, l = 0 (Given) The orbital is 1s.

(b) For n = 3 and l = 1 The orbital is 3p.

(c) For n = 4 and l = 2 The orbital is 4d.

(d) For n = 4 and l = 3  The orbital is 4f.

Answer:

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c) The given set of quantum numbers is not possible.
For a given value of n, ‘l ’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.

(d) The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible. For n = 3,
l = 0 to (3 – 1)
l = 0 to 2 i.e., 0, 1, 2

(f) The given set of quantum numbers is possible.

Answer:

(a) Total number of electrons in an atom for a value of n = 2n²
For n = 4,
Total number of electrons = 2 (4)² = 32
The given element has a fully filled orbital as
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰.
Hence, all the electrons are paired.
∴ Number of electrons (having n = 4 and m_s = – ½ ) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

Answer:

Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

\(mvr\)=\(n\frac{h}{2π}\) ….. (1)

Where, n = 1, 2, 3, …
According to de Broglie’s equation: λ = h/mv

or mv = h/ λ …….. (2)

Substituting the value of ‘mv’ from expression (2) in expression (1):

hr/λ = \(n\frac{h}{2π}\)

2πr = nλ ……. (3)

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Answer:

For He+ ion, the wave number \(\bar v\) associated with the Balmer transition, n = 4 to n

\(\bar v\) = 1/λ =RZ²\((\frac{1}{n_1^2}-\frac{1}{n_2^2})\)

Where, n₁ = 2 n₂ = 4
Z = atomic number of helium

\(\bar v\) = 1/λ = R(2)²\((\frac{1}{2^2}-\frac{1}{4^2})\)

= 4R(\(\frac{4-1}{16}\))

\(\bar v\) = 1/λ = \(\frac{3R}{4}\)

λ = \(\frac{4}{3R}\)

According to the question, the desired transition for hydrogen will have the same wavelength as that of He⁺.

R(1)² \((\frac{1}{n_1^2}-\frac{1}{n_2^2})\) = \(\frac{3R}{4}\)

\((\frac{1}{n_1^2}-\frac{1}{n_2^2})\) = \(\frac{3}{4}\) ……. (1)

By hit and trail method, the equality given by equation (1) is true only when
n₁ = 1 and n₂ = 2.
The transition for n₂ = 2  to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He⁺ spectrum.

Answer:

Energy associated with hydrogen-like species is given by,

E_n= – 2.18 × 10^–18 \(\frac{Z^2}{n^2}\) J

For ground state of hydrogen atom,

ΔE = E_∞ -E₁

= 0 – (- 2.18 × 10^–18 \(\frac{1^2}{1^2}\) J)

ΔE = 2.18 × 10^–18 J

For the given process, He⁺_(g) → He²⁺_(g) + e^–

An electron is removed from n = 1 to n = ∞.

ΔE = E_∞ -E₁

 = 0 – (- 2.18 × 10^–18 \(\frac{2^2}{1^2}\) J)

= 8.72 × 10^–18 J

∴ The energy required for the process = 8.72 × 10^–18 J

Answer:

1 m = 100 cm
1 cm = 10^–2 m
Length of the scale = 20 cm
= 20 × 10^–2 m
Diameter of a carbon atom = 0.15 nm
= 0.15 × 10^–9 m
One carbon atom occupies 0.15 × 10^–9 m.
∴ Number of carbon atoms that can be placed in a straight line

=  \(\frac{20 × 10^{-2}}{0.15 × 10^{-9}}\)

= 133.33 × 10⁷

= 1..33 × 10⁹

Answer:

Length of the given arrangement = 2.4 cm
Number of carbon atoms present = 2 × 10⁸
∴ Diameter of carbon atom = ( 2.4 × 10^-2 m) / (2 × 10⁸ )

= 1.2 × 10^-10 m

∴ Radius of carbon atom = Diameter of carbon atom / 2 = 6.0 × 10^-11 m

Answer:

(a)
Radius of zinc atom = Diameter / 2

= 2.6Å / 2

= 1.3 × 10^-10 m

= 130 × 10^-12 m

= 130 pm

(b) Length of the arrangement = 1.6 cm
= 1.6 × 10^–2 m
Diameter of zinc atom = 2.6 × 10^–10 m
∴ Number of zinc atoms present in the arrangement = (1.6 × 10^–2 m) / ( 2.6 × 10^–10 m)

= 6.153 × 10⁷ 

Answer:

Charge on one electron = 1.6022 × 10^–19 C
⇒ 1.6022 × 10^–19C charge is carried by 1 electron.
∴ Number of electrons carrying a charge of 2.5 × 10^–16 C 

=  (2.5 × 10^–16 C )/ (1.6022 × 10^–19 C)

= 1560 

Answer:

Charge on the oil drop = 1.282 ×10^–18C
Charge on one electron = 1.6022 × 10^–19C
Number of electrons present on the oil drop

= 1.282 ×10^–18C / 1.6022 × 10^–19C

= 8.0

Answer:

A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.
Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of α-particles (positively charged).

Answer:

The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is \(_{Z}^{A}X\)

Hence, \(_{35}^{79}Br\) is acceptable but \(_{79}^{35}Br\) is not acceptable.

\(^{79}Br\)  can be written but  \(^{35}Br\)  cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes.

Hence, it is necessary to mention the atomic mass of an element.

Answer:

Let the number of protons in the element be x.
∴ Number of neutrons in the element
= x + 31.7% of x
= x + 0.317 x
= 1.317 x
According to the question,
Mass number of the element = 81
∴ (Number of protons + number of neutrons) = 81

x + 1.317 x = 81

x = 81 /2.317

x = 35

Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
∴  The atomic symbol of the element is \(_{35}^{81}Br\)

Answer:

Let the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present
= x + 11.1% of x
= x + 0.111 x
= 1.111 x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
∴ Number of protons in the neutral atom = x – 1
Given,
Mass number of the ion = 37
∴ (x – 1) + 1.111x = 37
2.111x = 38

x = 18

Number of protons in the neutral atom = x – 1 = 17
∴ The symbol of the ion is \(_{17}^{37}Cr\)

Answer:

Let the number of electrons present in ion A³⁺ be x.

∴ Number of neutrons in it = x + 30.4% of x = 1.304 x
Since the ion is tri-positive,
⇒ Number of electrons in neutral atom = x + 3
∴ Number of protons in neutral atom = x + 3

Given,

Mass number of the ion = 56

∴ (x + 3) + 1.304 x = 56

2.304 x = 53

∴ x =  23

∴ Number of protons = x + 3 = 23 + 3 = 26

∴  The symbol of the ion \(_{26}^{56}Fe^{3+}\)

Answer:

The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays
The increasing order of wavelength is as follows:
Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

Answer:

Power of laser = Energy with which it emits photons
Power = E = \(Nhc \over λ\)

Where,

N = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of Energy (E):

= \((5.6 × 10^{24})(6.626 × 10^{-34} Js)(3 × 10^8 ms^{-1}) \over {(337.1  × 10^{-9}m)}\)

= 0.3302 × 10⁷ J
= 3.33 × 10⁶ J
Hence, the power of the laser is 3.33 × 10⁶ J.

Answer:

Wavelength of radiation emitted = 616 nm = 616 × 10^–9 m (Given)
(a) Frequency of emission ν 

ν = \(c\over λ\)

Where, c = velocity of radiation λ = wavelength of radiation
Substituting the values in the given expression of  ν

ν = \(3.0 × 10^8 ms^{-1} \over {616 × 10^{-9} m}\)

= 4.87 × 10⁸ × 10⁹ × 10^–3 s^–1 ν

= 4.87 × 10¹⁴ s^–1

Frequency of emission (ν) = 4.87 × 10¹⁴ s^–1
(b) Velocity of radiation, (c) = 3.0 × 10⁸ ms^–1
Distance travelled by this radiation in 30 s
= (3.0 × 10⁸ ms^–1) (30 s)
= 9.0 × 10⁹ m

(c) Energy of quantum (E) = hν
(6.626 × 10^–34 Js) (4.87 × 10¹⁴ s^–1)
Energy of quantum (E) = 32.27 × 10^–20 J

(d) Energy of one photon (quantum) = 32.27 × 10^–20 J
Therefore, 32.27 × 10^–20 J of energy is present in 1 quantum.
Number of quanta in 2 J of energy

= \( 2 J\over {32.27 × 10^{-20} J}\)

= 6.19 ×10¹⁸
= 6.2 ×10¹⁸

Answer:

From the expression of energy of one photon (E),

= \(hc\over λ\)

Where, λ = wavelength of radiation

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of E:

E = \((6.626 × 10^{-34} Js)(3.0 × 10^8 ms^{-1}) \over {699 × 10^{-9} m}\)

= 3.313 × 10^–19 J

Energy of one photon = 3.313 × 10^–19 J
Number of photons received with 3.15 × 10^–18 J energy

= 3.15 × 10^–18 J / 3.313 × 10^–19 J

= = 9.5
≈ 10

Answer:

Frequency of radiation (ν),

ν = \(1\over {2.0 × 10^{-9} s}\)

ν = 5.0 × 10⁸ s^–1
Energy (E) of source = Nhν
Where,
N = number of photons emitted,  h = Planck’s constantm  ν = frequency of radiation
Substituting the values in the given expression of (E):

E = (2.5 × 10¹⁵) (6.626 × 10^–34 Js) (5.0 × 10⁸ s^–1)

E = 8.282 × 10^–10 J

Hence, the energy of the source (E) is 8.282 × 10^–10 J.

Answer:

(i) Given,
Wavelength associated with first transition, λ₁ =589 nm =589 × 10^–9m
Wavelength associated with second transition, λ₂ = 589.6 nm =589.6 × 10^–9 m
Frequency of first wavelength is v₁ =

\(c\over λ_1\)

= \((3.0 × 10^8 ms^{-1}) \over {589 × 10^{-9} m}\)

= 5.093 × 10¹⁴ s^–1

And, frequency of second wavelength is v₂

v₂ =\(c\over λ_2\) = \((3.0 × 10^8 ms^{-1}) \over {589.6 × 10^{-9} m}\)

= 5.088 × 1014s-1

(ii) Energy difference between two excited states is given as,
ΔE = hv₁ – hv₂
Or, ΔE = h(v₁ – v₂)
= 6.626 × 10^-34 Js × (5.093 × 10¹⁴ s^-1 – 5.088 × 10¹⁴ s^-1)

= = 6.626 × 10^-34 Js × 0.005 × 1014 s^-1
= 3.31 × 10^-22 J

Answer:

It is given that the work function (W₀) for caesium atom is 1.9 eV.

(a) From the W₀ = \(hc \over λ_0\)

_λ0 = \(hc \over W_0\)

Where, _λ0 = threshold wavelength h = Planck’s constant c = velocity of radiation
Substituting the values in the given expression of (_λ0):

_λ0 = \((6.626 × 10^{-34} Js)(3 × 10^8 ms^{-1}) \over {(1.9 × 1.602  × 10^{-19} J)}\)

_λ0 = 6.53 × 10^–7 m

Hence, the threshold wavelength _λ0 is 653 nm.

(b) From the expression, W₀ = hν₀, we get:

ν₀ = \(W_0 \over h\)

Where, ν₀ = threshold frequency h = Planck’s constant
Substituting the values in the given expression of ν₀:

= \((1.9 × 1.602  × 10^{-19} J) \over {(6.626 × 10^{-34} Js)}\) …… (1 eV = 1.602 × 10^–19 J)

ν₀ = 4.593 × 10¹⁴ s^–1
Hence, the threshold frequency of radiation (ν₀) is 4.593 × 10¹⁴ s^–1

(c) According to the question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ν₀)

= hc (\(\frac{1}{λ} – \frac{1}{λ_0}\) )

= 6.626 × 10^-34 × 3 × 10⁸  ( \(λ-λ_0 \over {λλ_0}\) )

= ( 6.626 × 10^-34 )( 3 × 10⁸) (\(((653-500)10^{-9}) \over {(653)(500)10^{-18}}\) )

= 9.3149 × 10^–20 J
Kinetic energy of the ejected photoelectron = 9.3149 × 10^–20 J

Since K.E = \(\frac{1}{2}mv^2\) = 9.3149 × 10^–20 J

ν = \(\sqrt{2 × (9.3149 × 10^{-20} J) \over (9.10939 × 10^{-31}kg)}\)

= \(\sqrt{2.0451 × 10^{11} m^2 s^{-2}} \)

v = 4.52 × 10⁵ ms^–1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 10⁵ ms^–1.

Answer:

(a) Assuming the threshold wavelength to be λ₀ (nm) = λ₀ × 10^-9 m the kinetic energy of the radiation is given as:
Since K.E

h(v-v₀) = \(\frac{1}{2}mv^2\)

Three different equalities can be formed by the given value as:

hc (\(\frac{1}{λ} – \frac{1}{λ_0}\) ) = \(\frac{1}{2}mv^2\)

hc (\(\frac{1}{500 × 10^{-9}} – \frac{1}{λ_0 × 10^{-9}}\) ) = \(\frac{1}{2}m(2.55 × 10^5 × 10^{-2}) \)

\(\frac{hc}{10^{-9}}\)(\(\frac{1}{500 } – \frac{1}{λ_0 }\) ) = \(\frac{1}{2}m(2.55 × 10^3)\) ….(1)

Similarly,

\(\frac{hc}{10^{-9}}\)(\(\frac{1}{450 } – \frac{1}{λ_0 }\) ) = \(\frac{1}{2}m(3.45 × 10^3)\) ….(2)

\(\frac{hc}{10^{-9}}\)(\(\frac{1}{400 } – \frac{1}{λ_0 }\) ) = \(\frac{1}{2}m(5.35 × 10^3)\) ….(3)

Dividing equation (3) by equation (1):

So, threshold wavelength λ₀ = 540 nm

Note: part (b) of the question is not done due to the incorrect values of velocity given in the question.

Answer:

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = W0 + K.E
⇒ W0 = E – K.E
Energy of incident photon (E) = \(hc \over λ\)

Where, c = velocity of radiation
h = Planck’s constant
λ = wavelength of radiation
Substituting the values in the given expression of E:

E = 4.83 eV

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence, K.E = 0.35 V

K.E = 0.35 eV

∴ Work function, W₀ = E – K.E

= 4.83 eV – 0.35 eV
= 4.48 eV

Answer:

Energy of incident photon (E) is given by,

E = \(hc\over λ\)

Energy of the electron ejected (K.E)

= 10.2480 × 10^–17 J
= 1.025 × 10^–16 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
= 13.252 × 10^–16 J – 1.025 × 10^–16 J
= 12.227 × 10^–16 J

Answer:

Wavelength of transition = 1285 nm
= 1285 × 10^–9 m (Given)

Hence, for the transition to be observed at 1285 nm, n = 5.
The spectrum lies in the infra-red region.

Answer:

The radius of the n^th orbit of hydrogen-like particles is given by,

For radius (r₁) = 1.3225 nm
= 1.32225 × 10^–9 m
= 1322.25 × 10^–12 m
= 1322.25 pm

Similarly,

⇒ n₁ = 5 and n₂ = 2
Thus, the transition is from the 5^th orbit to the 2^nd orbit. It belongs to the Balmer series.
wave number \(\bar v\) for the transition is given by,

= 1.097 × 10⁷  (\(\frac {1}{2^2} -\frac{1}{5^2}\))

= 1.097 × 10⁷ (\(\frac {21}{100}\) )

= 2.303 × 10⁶ m^–1
Wavelength (λ) associated with the emission transition is given by,

λ = \(\frac {1}{\bar v}\)

= \(\frac {1}{2.303 × 10^6 m^{-1}}\)

= 0.434 ×10^–6 m λ

= 434 nm

Answer:

From de Broglie’s equation,

λ = \(\frac{h}{mv}\)

= \(\frac{6.626 × 10^{-34}Js}{(9.10939 × 10^{-31}kg)(1.6 × 10^6 ms^{-1}}\)

= 4.55 × 10^–10 m

λ = 455 pm

de Broglie’s wavelength associated with the electron is 455 pm.

Answer:

From de Broglie’s equation,

λ = \(\frac{h}{mv}\)

v = \(\frac{h}{mλ}\)

Where, v = velocity of particle (neutron)

h = Planck’s constant

m = mass of particle (neutron)

λ = wavelength

Substituting the values in the expression of velocity (v),

v = \(\frac{6.626 × 10^{-34}Js}{(1.67493 × 10^{-27}kg)(800 × 10^{-12} m}\)

= 4.94 × 10² ms^–1

v = 494 ms^–1

∴ Velocity associated with the neutron = 494 ms^–1

Answer:

From de Broglie’s equation,

λ = \(\frac{h}{mv}\)

Where, v = velocity of particle 

h = Planck’s constant

m = mass of particle (neutron)

λ = wavelength

Substituting the values in the expression of λ:

λ = \(\frac{6.626 × 10^{-34}Js}{(9.10939 × 10^{-31}kg)(2.19 × 10^6 ms^{-1}}\)

= 332 × 10^-12 m

= 332 pm

∴Wavelength associated with the electron = 332 pm

Answer:

rom de Broglie’s equation,

λ = \(\frac{h}{mv}\)

Substituting the values in the expression,

λ = \(\frac{6.626 × 10^{-34}Js}{0.1kg)(4.37 × 10^5 ms^{-1}}\)

= 1.516  × 10^-38 m

Answer:

From Heisenberg’s uncertainty principle,

Δx × Δp = \(\frac{h}{4π}\) ⇒ Δp = \(\frac{h}{(Δx)(4π)}\)

Where,
Δx = uncertainty in position of the electron
Δp = uncertainty in momentum of the electron
Substituting the values in the expression of Δp:

Δp = \(\frac{6.626 × 10^{-34}Js}{(0.002 nm)(4 × 3.14)}\)

= \(\frac{6.626 × 10^{-34}Js}{(2 × 10{-12}m)(4 × 3.14)}\)

= 2.637 × 10^–23 Jsm^–1
Δp = 2.637 × 10^–23 kgms^–1 (1 J = 1 kgms²s^–1)

Uncertainty in the momentum of the electron = 2.637 × 10^–23 kgms^–1

Actual momentum = \(\frac{h}{(0.05nm)(4π_m)}\)

= \(\frac{6.626 × 10^{-34}Js}{(5 × 10{-11}m)(4 × 3.14)}\)

= 1.055 × 10^–24 kgms^–1

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

Answer:

For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

Answer:

Nuclear charge experienced by an electron (present in a multi-electron atom) is dependent upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases.

Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

Answer:

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.

(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.

(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

Answer:

Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

Answer:

(a) Phosphorus (P):
Atomic number = 15
The electronic configuration of P is:
1s² 2s² 2p⁶ 3s² 3p³
The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons.

(b) Silicon (Si):
Atomic number = 14

The electronic configuration of Si is:
1s² 2s² 2p⁶ 3s² 3p²
The orbital picture of P can be represented as:

From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr):
Atomic number = 24
The electronic configuration of Cr is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe):
Atomic number = 26
The electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
The orbital picture of chromium is:

From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr):
Atomic number = 36
The electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
The orbital picture of krypton is:

Since all orbitals are fully occupied, there are no unpaired electrons in krypton

Answer:

(a) n = 4 (Given)
For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).
∴ l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the n^th shell = n²
For n = 4
Number of orbitals = 16
If each orbital is taken fully, then it will have 1 electron with m_s  value of -1/2 

∴Number of electrons with m_s value of -1/2 = 16