A) 82.35%
B) 97.85%
C) 42.35%
D) 59.45%
Correct Answer: B
Solution :
Kinetic energy of photoelectron \[KE=500\,keV=500\times {{10}^{3}}eV\] \[KE=m{{c}^{2}}-{{m}_{0}}{{c}^{2}}\] \[\frac{KE}{{{m}_{0}}{{c}^{2}}}=\frac{m{{c}^{2}}-{{m}_{0}}{{c}^{2}}}{{{m}_{0}}}\] \[=\frac{m-{{m}_{0}}}{{{m}_{0}}}=\frac{\Delta m}{{{m}_{0}}}\] \[\frac{\Delta m}{m}=\frac{KE}{{{m}_{0}}{{c}^{2}}}\] Hence % increase in mass is \[=\frac{\Delta m}{m}\times 100=\frac{KE}{{{m}_{0}}{{c}^{2}}}\times 100\] \[=\frac{500\times {{10}^{3}}}{0.511\times {{10}^{6}}}\times 100\] \[=\frac{5}{5.11}\times 100=97.85%\]
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