Atoms And Nuclei Question 317

Question: A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be

Options:

A) $ 800\overset{o}{\mathop{A}}, $

B) $ 825\overset{o}{\mathop{A}}, $

C) $ 975\overset{o}{\mathop{A}}, $

D) $ 01025\overset{o}{\mathop{A}}, $

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Answer:

Correct Answer: C

Solution:

  • Number of possible spectral lines emitted when an electron jumps back to ground state from nth orbit $ \frac{n( n-1 )}{2} $ Here, $ \frac{n( n-1 )}{2}=6\Rightarrow n=4 $ Wavelength $ \lambda $ from transition from n = 1 to n = 4 is given by, $ \frac{1}{\lambda }=R( \frac{1}{1}-\frac{1}{4^{2}} )\Rightarrow \lambda =\frac{16}{15R}=975\overset{o}{\mathop{A}}, $


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