A) Proton
B) Deutron
C) Both have equal values
D) None of these
Correct Answer: A
Solution :
[a] de-Broglie wavelength of a charged particle is given by, \[\lambda \propto \frac{1}{\sqrt{mq}}\] If \[{{m}_{p}}\] and e are mass and charge of a proton respectively, and, \[{{m}_{D}}\] and e are mass and charge of a deuteron respectively, then \[\frac{{{\lambda }_{p}}}{{{\lambda }_{D}}}=\sqrt{\frac{{{m}_{D}}{{q}_{D}}}{{{m}_{p}}{{q}_{p}}}}=\sqrt{\frac{\left( 2{{m}_{p}} \right)\left( e \right)}{\left( {{m}_{p}} \right)\left( e \right)}}=\sqrt{2}\] \[{{\lambda }_{p}}=\sqrt{2}{{\lambda }_{D}}\] Momentum, is given by, \[P=\frac{h}{\lambda }\text{ or, p}\propto \frac{1}{\lambda }\] Where, h = plank's constant Since the wavelength of a proton is more than that of deuteron thus, the momentum of a proton is lesser than that of deuteron. Hence, the momentum of proton is less.
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