atoms Question 26
Question: Q. 4. (i) Using Bohr’s second postulate of quantization of orbital angular momentum, show that the circumference of the electron in the $n^{\text {th }}$ orbital state in hydrogen atom is $n$ times the de-Broglie wavelength associated with it.
(ii) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?
U A [Delhi I, II, III 2012]
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Solution:
Ans. (i) According to Bohr’s second postulate
$$ \begin{aligned} m v r_{n} & =\frac{n h}{2 \pi} \ \Rightarrow \quad 2 \pi r_{n} & =\frac{n h}{m v} \end{aligned} $$
$$ \begin{align*} \text { But } & & \frac{h}{m v} & =\frac{h}{p}=\lambda \ & \therefore & 2 \pi r_{n} & =n \lambda \end{align*} $$
[Note : If the student just writes $m v r_{n}=\frac{n h}{2 \pi}$ and writes $\lambda=\frac{h}{p}$ award 2 marks]
(ii) For third excited state, $n=4$
For ground state, $\quad n=1$
Hence, possible transitions are
$$ \begin{aligned} & n_{f}=4 \text { to } n_{i}=3,2,1 \ & n_{f}=3 \text { to } n_{i}=2,1 \ & n_{f}=2 \text { to } n_{i}=1 \end{aligned} $$
Total number of transitions $=6$
Note : If transition is taking place nth orbit to 1st orbit, number of possible spectral lines are $\frac{n(n-1)}{2}$.
[CBSE Marking Scheme 2012]
Commonly Made Error
- Some students are unable to recall the correct equations
$$ \text { i.e., } m v r_{n}=\frac{n h}{2 \pi} \text { and } \lambda=\frac{h}{p} $$
