The de Broglie wavelength of an electron moving with a velocity $1.5 × 10^8 m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electro to the energy of the photon is:
Answer & explanation
Correct answer: option 1
de Broglie wavelength of an electron, $λ_e=\frac{h}{m_ev_e}$
where h is the Plank’s constant, $m_e$ and $v_e$ is the mass and velocity of the electron respectively.
Wavelength of a photon = $λ_{ph}$
As per question
$λ_{ph}=λ_e=\frac{h}{m_ev_e}$ (i)
Kinetic energy of the electron,
$K_e=\frac{1}{2}m_ev_e^2$ (ii)
Energy of the photon,
$E_{ph}=\frac{hc}{λ_{ph}}$ (iii)
Divide (ii) and (iii), we get
$\frac{K_e}{E_{ph}}=\frac{\frac{1}{2}m_ev_e^2}{\frac{hc}{λ_{ph}}}=\frac{1}{2}\frac{m_ev_e^2λ_{ph}}{hc}=\frac{1}{2}\frac{m_ev_e^2h}{hcm_ev_e}$ (Using (i))
$=\frac{1}{2}\frac{v_e}{c}=\frac{1.5 × 10^8 m/s}{2×3×10^8 m/s}=\frac{1}{4}$