Question

Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..

• A. 1.16 eV
• B. 2.38 eV
• C. 0.58 eV
• D. 2.98 ev

Answer 1

The correct option is C 0.58 eV

Work function of plate $\varphi =1.9eV$
Wavelength of incident photon $\lambda =5000A^o=500nm$
Energy of the incident photon $E_{incident}=\frac{1240}{\lambda \left(innm\right)}eV$
$⟹E_{incident}=\frac{1240}{500}=2.48eV$
Maximum kinetic energy of photon emitted $K.E_{max}=E_{incident}-\varphi$
$⟹K.E_{max}=2.48-1.9=0.58eV$