Question

Monochromatic radiation of wavelength 640.2 nm (1nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Answer 1

from photoelectric effect,

$E=\frac{hc}{\lambda}=W_0+\frac{1}{2}mv_{max}^2$

we also know, maximum kinetic energy , $\frac{1}{2}mv_{max}^2=eV_0$, where $V_0$ is the stopping potential.

given, $\lambda=640.2nm$, $V_0$=0.54V

so, (6.63 × 10^-34 × 3 × 10^8)/(640.2 × 10^-9 × 1.6 × 10^-19) eV = $W_0$ + 0.54eV

or, 1.94 eV = $W_0$ + 0.54eV

or, $W_0$ = 1.4eV

Now the source is replaced by iron source which produce 427.2nm wavelength.

then, $E=\frac{hc}{\lambda}=W_0+\frac{1}{2}mv_{max}^2$

or, (6.63 × 10^-34 × 3 × 10^8)/(427.2 × 10^-9 × 1.6 × 10^-19)eV = 1.4eV + $K.E_{max}$

or, 2.9eV = 1.4eV + $K.E_{max}$

or, $K.E_{max}$ = 1.5eV = e$V_0$

so, stopping potential, $V_0$ = 1.5Volt