Question

When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.

Answer 1

Given:
Wavelength of light, $\lambda$ = 400 nm = 400$\times$10^$-9$ m
Stopping potential, V₀ = 1.1 V
From Einstein's photoelectric equation,
$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+e{V}_0$,
where h = Planck's constant
c= speed of light
$\lambda$ = wavelength of light
${\lambda }_0$ = threshold wavelength
$V_0$ = stopping potential
On substituting the respective values in the above formula, we get:
$$\begin{gathered} \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{{\lambda }_0}+1.6\times {10}^{-19}\times 1.1 \\ \Rightarrow 4.97\times {10}^{-19}=\frac{19.89\times {10}^{-26}}{{\lambda }_0}+1.76\times {10}^{-19} \\ \Rightarrow 4.97=\frac{19.89\times {10}^{-7}}{{\lambda }_0}+1.76 \\ \Rightarrow \frac{19.89\times {10}^{-7}}{{\lambda }_0}=4.97-1.76=3.21 \\ \Rightarrow {\lambda }_0=\frac{19.89\times {10}^{-7}}{3.21} \\ =6.196\times {10}^{-7}\mathrm{m}=620\mathrm{nm} \end{gathered}$$