CBSE Class 12 · 2019 · Modern PhysicsShort Answer

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State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum it belongs.

Answer: $L = nh /2 π$ ; $λ_{min} \approx 1458 nm$ (infrared).

Solution

Bohr's quantisation condition: the angular momentum of the orbiting electron is an integral multiple of

\frac{h}{2 π}

L = m v r = \frac{nh}{2 π}, n = 1, 2, 3, \dots

For the Brackett series

n_{f} = 4

; the shortest wavelength (series limit) corresponds to

n_{i} = \infty

\frac{1}{λ} = R (\frac{1}{4 ^{2}} - \frac{1}{\infty ^{2}}) = \frac{R}{16} = \frac{1.097 \times 1 0 ^{7}}{16} = 6.86 \times 1 0^{5} m^{- 1} .

λ = 1.458 \times 1 0^{- 6} m \approx 1458 nm (infrared) .

λ \approx 1458 nm (infrared)

CBSE Class 12 Board · 2019 · Subjective

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