The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be :
Solution
($\Delta$E) Releases when photon going from n = 5 to n = 1<br><br>$\Delta$E = (13.6 $-$ 0.54) eV = 13.06 eV.<br><br>P<sub>i</sub> = P<sub>f</sub> (By linear momentum conservation)<br><br>$$0 = {h \over \lambda } - Mv = {V_{{\mathop{\rm Re}\nolimits} coil}} = {h \over {\lambda M}}$$ ..... (i)<br><br>& $$\Delta E = {{hc} \over \lambda } = {{hc} \over {\lambda M}} \times M = Mc{V_{{\mathop{\rm Re}\nolimits} coil}}$$<br><br>$${V_{{\mathop{\rm Re}\nolimits} coil}} = {{\Delta E} \over {Mc}} = {{13.06 \times 1.6 \times {{10}^{ - 19}}} \over {1.67 \times {{10}^{ - 27}} \times 3 \times {{10}^8}}}$$ = 4.17 m/sec
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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