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} $
Show Answer
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} $
