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
$KE=500keV=500\times {10}^{3}eV$
$KE=m{c}^2-m_0{c}^2$
$\frac{KE}{m_0{c}^2}=\frac{m{c}^2-m_0{c}^2}{m_0}$
$=\frac{m-m_0}{m_0}=\frac{\Delta m}{m_0}$
$\frac{\Delta m}{m}=\frac{KE}{m_0{c}^2}$
Hence % increase in mass is
$=\frac{\Delta m}{m}\times 100=\frac{KE}{m_0{c}^2}\times 100$
$=\frac{500\times {10}^3}{0.511\times {10}^6}\times 100$
$=\frac{5}{5.11}\times 100=97.85$