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Q.

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

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a

$E_{1} < E_{2}$

b

information insufficient

c

$E_{1} = E_{2}$

d

$E_{1} > E_{2}$

answer is B.

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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 $E_{1} < E_{2}$

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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 E1towards right. Second one is moving towards left with same speed and emits a photon of energy E2 towards left. Taking recoil of nucleus into account during emission process