Atoms And Nuclei Question 36

Question: An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. The wave number of the emitted radiations (R = Rydberg’s constant) will be [CBSE PMT 1995]

Options:

A) $ \frac{16}{3R} $

B) $ \frac{2R}{16} $

C) $ \frac{3R}{16} $

D) $ \frac{4R}{16} $

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Answer:

Correct Answer: C

Solution:

Wave number $ \frac{1}{\lambda }=R[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ]=R[ \frac{1}{4}-\frac{1}{16} ]=\frac{3R}{16} $



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