CBSE Class 12 · 2012 · Modern PhysicsShort Answer

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(a) Using Bohr's second postulate of quantization of orbital angular momentum, show that the circumference of the electron in the

n^{th}

orbital state in a hydrogen atom is

n

times the de-Broglie wavelength associated with it.
(b) The electron in a hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines that can be emitted when it finally moves to the ground state?

Answer: $2 π r = nλ$ ; maximum $6$ spectral lines (from $n = 4$ ).

Solution

(a) Bohr's second postulate:

m v r = \frac{nh}{2 π}

. The de-Broglie wavelength is

λ = \frac{h}{m v}

, so

m v = \frac{h}{λ}

. Substituting:

\frac{h}{λ} r = \frac{nh}{2 π} \Rightarrow 2 π r = nλ .

Thus the circumference of the

n^{th}

orbit contains exactly

n

de-Broglie wavelengths (a standing wave).
(b) The third excited state is

n = 4

. Maximum number of spectral lines:

N = \frac{n ( n - 1 )}{2} = \frac{4 \times 3}{2} = 6.

2 π r = nλ; N = 6 lines

CBSE Class 12 Board · 2012 · Subjective

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CBSE Class 12 · 2012 · Modern PhysicsShort Answer

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