On shining light of wavelength `6.2xx10^(-6)` m on a metal surface photo-electrons are emitted. The work function of the metal is 0.1 eV. Find the kinetic energy of a photo-electron (in eV)

Options

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer

A

Explanation

Here, `lambda=6.2xx10^(-6) m , phi_0`= 0.1 eV
Energy of the incident photon , `E=hupsilon=(hc)/lambda`
`E=((6.6xx10^(-34))(3xx10^8))/(6.2xx10^(-6))`J
`=(6.6xx3xx10^(-26))/((6.2xx10^(-6))(1.6xx10^(-19)))` eV=0.2 eV
As `E=K+phi_0 , K=E- phi_0`=0.2 eV - 0.1 eV = 0.1 eV