Atoms And Nuclei Question 110
Question: An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as $ 10^{7} $ per metre. What will be the frequency of radiation emitted [Pb. PMT 2001]
Options:
A) $ 6.75\times 10^{12},Hz $
B) $ 6.75\times 10^{14},Hz $
C) $ 6.75\times 10^{13},Hz $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
By using $ \nu =RC,[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ] $
$ \Rightarrow \nu =10^{7}\times (3\times 10^{8}),[ \frac{1}{4^{2}}-\frac{1}{5^{2}} ] $ = 6.75 ´ 1013 Hz
