Question
In an experiment on photoelectric effect on a metal surface, one finds a stopping potential of 1.8 V for the wavelength of 300 nm and a stopping potential of 0.9 V for the wavelength of 400 nm. The cutoff wavelength λ0 (the maximum wavelength that can produce photoelectric effect) for the metal is
Options
Solution
[Key Concept: Einstein's equation eVs = hc(1/λ − 1/λ0); the ratio of two stopping potentials eliminates hc.]
from ${eV}_{0} = hC\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$
$$\begin{array}{r} 1.8eV = hC\left( \frac{1}{300\text{ }nm} - \frac{1}{\lambda_{0}} \right)\#(i) \\ 0.9eV = hC\left( \frac{1}{400\text{ }nm} - \frac{1}{\lambda_{0}} \right)\#(ii) \end{array}$$
from (i) & (ii)
$${\begin{matrix} & 2\left( \frac{1}{400\text{ }nm} - \frac{1}{\lambda_{0}} \right) = \frac{1}{300\text{ }nm} - \frac{1}{\lambda_{0}} \\ & \frac{1}{200\text{ }nm} - \frac{1}{300\text{ }nm} = \frac{2}{\lambda_{0}} - \frac{1}{\lambda_{0}} \\ & \frac{3 - 2}{600} = \frac{1}{\lambda_{0}} \end{matrix} }{\lambda_{0} = 600\text{ }nm }$$
Create a free account to view solution
View Solution Free