A hydrogen atom in the ground state absorbs 10.2 eV of energy. The orbital angular momentum…

Q: A hydrogen atom in the ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by

(a) 1.05 × 10-34 Js

(b) 2.11 × 10-34 Js

(c) 3.16 × 10-34 Js

(d) 4.22 × 10-34 Js

Ans: (a)
Sol:

Angular momentum $ \displaystyle L =\frac{n h}{2 \pi} $

here, n = 1

$ \displaystyle L =\frac{1\times 6.6 \times 10^{-34} }{2 \times 3.14} $

= 1.05 × 10-34 Js

The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å…

Q: The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å. The wavelength of the first line is

(a) $ \displaystyle \frac{27}{20}\times 4861 A^o $

(b) $ \displaystyle \frac{20}{27}\times 4861 A^o $

(c) $ \displaystyle 20 \times 4861 A^o $

(d) $ \displaystyle 4861 A^o $

Ans: (a)

Sol:

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{1}{2^2}-\frac{1}{4^2})$

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{1}{4}-\frac{1}{16})$

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{3}{16}) $ …(i)

$ \displaystyle \frac{1}{\lambda_1} = R (\frac{1}{2^2}-\frac{1}{3^2})$

$\displaystyle \frac{1}{\lambda_2} = R (\frac{1}{4}-\frac{1}{9})$

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{5}{36}) $ …(ii)

On dividing (i) by (ii)

$ \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{3R/16}{5R/36} $

$ \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{27}{20} $

$ \displaystyle \lambda_1 = \frac{27}{20}\times \lambda_2 $

$ \displaystyle \lambda_1 = \frac{27}{20}\times 4861 A^o $

The ratio of magnetic dipole moment to angular momentum in a hydrogen like atom…

Q: The ratio of magnetic dipole moment to angular momentum in a hydrogen like atom is

(a) e/m

(b) e/2m

(c) e/3m

(d) 2e/m

Ans: (b)

Sol:

Magnetic Moment , M = I A

$ \displaystyle M = \frac{e}{T}. A = \frac{e \omega}{2 \pi}.\pi r^2 $

$ \displaystyle M = \frac{1}{2}e \omega r^2 $

$ \displaystyle M = \frac{1}{2}e (\omega r )r $

$ \displaystyle M = \frac{1}{2}e v r $

$ \displaystyle M = \frac{1}{2 m}e ( m v r ) $

$ \displaystyle M = \frac{1}{2 m}e ( L ) $ (Where L = angular momentum )

$ \displaystyle \frac{M}{L} = \frac{e}{2 m} $

For certain atom, there are energy levels A, B, C corresponding to energy values EA < EB < EC...

Q: For certain atom, there are energy levels A, B, C corresponding to energy values EA < EB < EC. Choose the correct option if λ1 , λ2, λ3 are the wavelength of radiations corresponding to the transition from C to B, B to A and C to A respectively:

(a) λ3 = λ1 + λ2

(b) $ \displaystyle \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$

(c) λ1 + λ2 + λ3 = 0

(d) 3 λ2 = λ3 + 2 λ2

Ans: (b)

Sol:

$ \displaystyle \Delta E = h \nu = \frac{h c}{\lambda} $

$ \displaystyle E_C – E_A = ( E_C – E_B ) + ( E_B – E_A ) $

$ \displaystyle \frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} $

$ \displaystyle \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} $

$ \displaystyle \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} $

An electron revolves round a nucleus of atomic number Z. If 32.4 eV of energy is required to …

Q. An electron revolves round a nucleus of atomic number Z. If 32.4 eV of energy is required to excite an electron from the n = 3 state to n = 4 state, then the value of Z is

(a) 5

(b) 6

(c)4

(d) 7

Ans: (d)

Sol:E3 = −13.6Z2/32    and E4 = −13.6Z2/42

ΔE = E4 − E3

ΔE = ( −13.6Z2/42)-( −13.6Z2/32)

ΔE = 13.6Z2[1/32 – 1/42]

ΔE = 13.6Z2[7/9×16]

32.4 = 13.6Z2[7/9×16]

Z2 = (32.4 × 9 × 16)/(13.6×7)

Z2 = 49

Z = 7