A) 82.35%
B) 97.85%
C) 42.35%
D) 59.45%
Correct Answer: B
Solution :
Kinetic energy of photoelectron \[KE=500\text{ }keV=500\times {{10}^{3}}eV\] But \[KE=m{{c}^{2}}-{{m}_{0}}{{c}^{2}}\] \[\Rightarrow \] \[\frac{KE}{{{m}_{0}}{{c}^{2}}}=\left( \frac{m{{c}^{2}}-{{m}_{0}}{{c}^{2}}}{{{m}_{0}}{{c}^{2}}} \right)\] \[=\frac{m-{{m}_{0}}}{{{m}_{0}}}=\frac{\Delta m}{{{m}_{0}}}\] Or \[\frac{\Delta m}{m}=\frac{KE}{{{m}_{0}}{{c}^{2}}}\] Hence, % increase in mass is \[=\frac{\Delta m}{{{m}_{0}}}\times 100\] \[=\frac{KE}{{{m}_{0}}{{c}^{2}}}\times 100\] \[=\frac{500\times {{10}^{3}}eV}{0.511\times {{10}^{6}}eV}\times 100\] \[(\because {{m}_{0}}{{c}^{2}}=0.511\times {{10}^{6}}eV)\] \[=\frac{5}{5.11}\times 100=97.85%\]
You need to login to perform this action.
You will be redirected in
3 sec
