If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2 , emitted radiation has a frequency (R = Rydberg's constant, c = velocity of light)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{5 R c}{36}$

$\frac{R c}{25}$

$\frac{3 R c}{27}$

$\frac{8 R c}{9}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

When an electron jumps from orbit n_i to orbit n_f (n_i > n_f), the frequency of emitted photon is given by

$v = c R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$

$\Rightarrow v = c R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$

$v = c R (\frac{5}{36}) \Rightarrow v = \frac{5 c R}{36}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE