Two hydrogen atoms are in excited state with electrons residing in n = 2. First one is moving towards left and emits a photon of energy $E_{1}$ towards right. Second one is moving towards left with same speed and emits a photon of energy $E_{2}$ towards left. Taking recoil of nucleus into account during emission process

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$E_{1} > E_{2}$

$E_{1} < E_{2}$

$E_{1} = E_{2}$

information insufficient

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Energy released + Initial energy of hydrogen atom = Final energy of hydrogen atom + energy of photon .

In the first case K.E. of H-atom increases due to recoil whereas in the second case K.E. decreases due to recoil.

Therefore $E_{1} < E_{2}$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

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

Take Free Test