### NEET Solved Paper 2014 Question 41

##### Question: Hydrogen atom in ground state is excited by a monochromatic radiation of $ \lambda =975\overset{o}{\mathop{A}}, $ . Number of spectral lines in the resulting spectrum emitted will be [AIPMT 2014]

#### Options:

A) 3

B) 2

C) 6

D) 10

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

Energy provided to the ground state electron

$ =\frac{hc}{\lambda }=\frac{6.6\times {10^{-34}}\times 3\times 10^{8}}{975\times {10^{-10}}} $ $ =\frac{6.6\times 3}{975}\times {10^{-16}} $ $ =0.020\times {10^{-16}}=2\times {10^{-18}}J $

$ =\frac{20\times {10^{-19}}}{1.6\times {10^{-19}}}eV $ $ =\frac{20}{1.6}eV=12.5eV $

It means the electron is jump to n = 3 from n = 1.

The number of field lines possible from $ n=3 $ To $ n=1 $ is 2. n $\rightarrow$ 3, to n $\rightarrow$ 2 to n $\rightarrow$ 1.