Question

Light of wavelength 4000 A${}^o$ is incident on a metal plate whose work function is 20eV. The maximum kinetic energy of the emitted photoelectrons would be :

• A. 0.5 eV
• B. 1.1 eV
• C. 1.5 eV
• D. 2.0 eV

Answer 1

The correct option is B 1.1 eV

Wavelength of light, $\lambda =4000A^o=400nm$

Energy of incident photon $E=\frac{hc}{\lambda }$

Or $E=\frac{1240}{\lambda \left(innm\right)}eV=\frac{1240}{400}=3.1eV$

Maximum kinetic energy of photoelectrons $K.E_{max}=E-\varphi$

where $\varphi =2eV$

$⟹K.E_{max}=3.1-2=1.1eV$