Q84 / 100 · BITSAT 2022 · Physics

The de-Broglie wavelength of an electron moving with a velocity $\frac{c}{3}\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$ is equal to the wavelength of photon. The ratio of the kinetic energies of electron and photon is

  • A
    $1: 4$
  • Correct answer
    B
    $1: 3$
  • C
    $1: 2$
  • D
    $2: 1$

Solution

The de-Broglie wavelength, $\lambda=\frac{h}{m v}$ Here, $\quad \lambda_e=\frac{h}{m_e \frac{c}{3}}$ and $\lambda_p=\frac{h}{m_p c}$ Given, $\quad \lambda_e=\lambda_p$ So, $\quad \frac{h}{m_e \frac{c}{3}}=\frac{h}{m_p c}$ $\Rightarrow \quad \frac{m_e}{m_p}=3$ Ratio of KE $=\frac{K_e}{K_p}=\frac{\frac{1}{2} m_e v_e^2}{\frac{1}{2} m_p v_p^2}=\frac{1}{3}$ $\left(\because \frac{m_e}{m_p}=3, v_e=\frac{c}{3}\right.$ and $\left.v_p=c\right)$

Hence, the correct answer is option: 'B'.

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