JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Modern Physics

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Li^{++}

, electron in first Bohr orbit is excited to a level by a radiation of wavelength

λ

. When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of

λ

? (Given:

h = 6.63 \times 1 0^{34} js

;

e = 3 \times 1 0^{8} ms^{- 1}

)

(A)10.8 nm✓ Correct
(B)11.4 nm
(C)9.4 nm
(D)12.3 nm

Answer: (A)

Solution

Seeing six spectral lines on the way down means the ion was lifted to

n = 4

, because

(2 4) = 6

downward transitions are then possible. For

Li^{++}

with

Z = 3

, the energy absorbed from

n = 1

n = 4

E = 13.6 Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{4 ^{2}}) = 13.6 \times 9 \times \frac{15}{16} = 114.75 eV

. The wavelength that carries this energy is

λ = \frac{h c}{E} = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{114.75 \times 1.6 \times 1 0 ^{- 19}} \approx 10.8 nm

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Modern Physics

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