Atoms And Nuclei Question 20

Question: An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg’s constant $ R=10^{5}c{{m}^{-1}} $ . The frequency in Hz of the emitted radiation will be [CPMT 1976]

Options:

A) $ \frac{3}{16}\times 10^{5} $

B) $ \frac{3}{16}\times 10^{15} $

C) $ \frac{9}{16}\times 10^{15} $

D) $ \frac{3}{4}\times 10^{15} $

Show Answer

Answer:

Correct Answer: C

Solution:

$ \frac{1}{\lambda }=R,( \frac{1}{2^{2}}-\frac{1}{4^{2}} )=\frac{3R}{16}\Rightarrow \lambda =\frac{16}{3R}=\frac{16}{3}\times {{10}^{-5}}cm $

Frequency $ n=\frac{c}{\lambda }=\frac{3\times 10^{10}}{\frac{16}{3}\times {{10}^{-5}}}=\frac{9}{16}\times 10^{15}Hz $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक