Question

When electron is accelerated between 500 keV, what is the percentage increase in mass?

• A. 82.35%
• B. 97.85%
• C. 42.35%
• D. 59.45%

Answer 1

Answer: (B)

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% $