Question

A monochromatic light illuminates the surface of sodium and the maximum kinetic energy of a photoelectron is 1.2eV.The cutoff wavelength of sodium is 540nm, the wavelength of the incident light is.........

Answer 1

Given,

The kinetic energy of the photoelectric = 1.2eV

Cut-off wavelength of sodium $(\lambda_o)=540nm = 540\times 10^{-9}m$

Let the wavelength of the incident light be $\lambda.$

Now, applying the photoelectric formula,

$eV = \frac{hc}{\lambda}-\frac{hc}{\lambda_o}$

Now, substituting the values,

$\Rightarrow 1.2eV = \frac{1240 eV -nm}{\lambda}-\frac{1240eV-nm}{540nm}$

Now, substituting the values,

$\Rightarrow \frac{1.2eV}{1240 eV-nm}=\frac{1}{\lambda}-\frac{1}{540nm}$

$\Rightarrow \frac{1}{\lambda}=\frac{1.2}{1240nm}+\frac{1}{540nm}$

$\Rightarrow \frac{1}{\lambda}=9.68\times 10^{-4}+18.51\times10^{-4}$

$\Rightarrow \frac{1}{\lambda}=28.19\times 10^{-4}$

$\Rightarrow \lambda = 354.73nm$