JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 7)
The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be :
4.34 m/s
2.19 m/s
3.25 m/s
4.17 m/s
Explanation
($$\Delta$$E) Releases when photon going from n = 5 to n = 1
$$\Delta$$E = (13.6 $$-$$ 0.54) eV = 13.06 eV.
Pi = Pf (By linear momentum conservation)
$$0 = {h \over \lambda } - Mv = {V_{{\mathop{\rm Re}\nolimits} coil}} = {h \over {\lambda M}}$$ ..... (i)
& $$\Delta E = {{hc} \over \lambda } = {{hc} \over {\lambda M}} \times M = Mc{V_{{\mathop{\rm Re}\nolimits} coil}}$$
$${V_{{\mathop{\rm Re}\nolimits} coil}} = {{\Delta E} \over {Mc}} = {{13.06 \times 1.6 \times {{10}^{ - 19}}} \over {1.67 \times {{10}^{ - 27}} \times 3 \times {{10}^8}}}$$ = 4.17 m/sec
$$\Delta$$E = (13.6 $$-$$ 0.54) eV = 13.06 eV.
Pi = Pf (By linear momentum conservation)
$$0 = {h \over \lambda } - Mv = {V_{{\mathop{\rm Re}\nolimits} coil}} = {h \over {\lambda M}}$$ ..... (i)
& $$\Delta E = {{hc} \over \lambda } = {{hc} \over {\lambda M}} \times M = Mc{V_{{\mathop{\rm Re}\nolimits} coil}}$$
$${V_{{\mathop{\rm Re}\nolimits} coil}} = {{\Delta E} \over {Mc}} = {{13.06 \times 1.6 \times {{10}^{ - 19}}} \over {1.67 \times {{10}^{ - 27}} \times 3 \times {{10}^8}}}$$ = 4.17 m/sec
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