Question 31: Dual Nature of Matter and Radiation

Question

$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1=2 \lambda_2$ then the work function of material is given by :

Accepted Answer

According to the photoelectric effect, when light of a certain frequency or wavelength hits a metal surface, it provides energy to the electrons. Part of this energy is used to overcome the attractive forces of the metal (the work function, $\phi$ ), and the remaining energy becomes the maximum kinetic energy (K) of the emitted photoelectron. The energy of an incident photon is $E=\frac{h c}{\lambda}$. The equation is:
$ \mathrm{E}=\phi+\mathrm{K}_{\max } $
$\Rightarrow $ $\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{K}_{\max }$
$\Rightarrow $ $ \mathrm{K}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi $
We are given two different wavelengths and their corresponding kinetic energies:
For wavelength $\lambda_1$ :
$ \begin{equation*} \mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\phi \ldots 	ag{i} \end{equation*} $
For wavelength $\lambda_2$ :
$ \begin{equation*} \mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\phi \ldots 	ag{ii} \end{equation*} $
It is given that $\lambda_1=2 \lambda_2$.
$ \begin{equation*} \mathrm{K}_1=\frac{\mathrm{hc}}{2 \lambda_2}-\phi \ldots 	ag{iii} \end{equation*} $
From equation (ii), we can find an expression for $\frac{\mathrm{hc}}{\lambda_2}$ :
$ \frac{\mathrm{hc}}{\lambda_2}=\mathrm{K}_2+\phi $
Now, substituting this into equation (iii) :
$ \mathrm{K}_1=\frac{1}{2}\left(\mathrm{~K}_2+\phiight)-\phi $
$\Rightarrow $ $2 \mathrm{~K}_1=\mathrm{K}_2+\phi-2 \phi$
$\Rightarrow $ $2 \mathrm{~K}_1=\mathrm{K}_2-\phi$
$\Rightarrow $ $ \phi=\mathrm{K}_2-2 \mathrm{~K}_1 $
Therefore, the work function of the material is $K_2-2 K_1$.
Hence, the correct option is (D).

Answer

$ K_2+2 K_1 $

Answer

$ 2 K_2-K_1 $

Answer

$ K_1-2 K_2 $

Answer

$ K_2-2 K_1 $

Dual Nature of Matter and Radiation

K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1=2 \lambda_2$ then the work function of material is given by :

Choose the correct answer: