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  • 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

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 right with same speed and emits a photon of energy E_2 towards right. Taking recoil of nucleus into account during emission process

Option: 1

E_1 > E_2


Option: 2

E_1 <E_2


Option: 3

E_1 =E_2


Option: 4

none 


Answers (1)

best_answer

 

If only conservative forces act on a system, total mechnical energy remains constant -

K+U=E\left ( constant \right )

\Delta K+\Delta U=0

\Delta K=-\Delta U

-

 

 

In the first case K.E. of H-atom increases due to recoil whereas in the second case K.E. decreases due to recoil but E_1 + KE_1 = E_2 + KE_2 \\\\ \therefore E_2 > E_1

Posted by

vishal kumar

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