Question

Photoelectric work function of a metal is 1eV. Light of wave length =3000 falls on it. The photo electron come out with velocity

Answer 1

Work function of metal surface $\varphi =1eV$
Wavelength of incident photon $\lambda =3000A^o=300nm$
Energy of incident photon $E=\frac{1240}{\lambda \left(innm\right)}eV=\frac{1240}{300}=4.13eV$
Using $\frac{1}{2}m{v}_{max}^2=E-\varphi$
$\therefore \frac{1}{2}m{v}_{max}^2=4.13-1=3.13eV$
$\therefore$ $\frac{1}{2}\times 9.1\times {10}^{-31}{v}_{max}^2=3.13\times 1.6\times {10}^{-19}$
$⟹v_{max}=1\times {10}^{6}m/s=1000000m/s$
Correct answer is option D.