The de Broglie wavelength of an electron moving with a velocity c/2 (c=velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

Options

A. `1:4`
B. `1:2`
C. `1:1`
D. `2:1`

Answer

A

Explanation

de Brogile wavelength of an electron,
`lambda_e = (h)/(pe) = (h)/(m_e (c/2))=(2h)/(m_ec)`.....(i)
where `m_e` is the mass of an electron Kinetic energy of an electron,
`K_e =1/2 m_e (C/2)^2 = 1/8 m_e C^2`
Kinetic energy of a photon, `K_(ph)=(hc)/(lambdaph)`
`because lambda_e = lambda_(ph) `(Given)
`therefore K_(ph) = (hc)/(lambda_e)= (hc)/((2h)/(m_eC))=(m_ec^2)/(2)` (Using(i))
`therefore (K_e)/(K_(ph))= 1/8 m_eC^2 xx (2)/(m_ec^2)=1/4`