Question

If light of wavelength $\lambda _{1}$ is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is $E_{1}$ . If wavelength of light changes to $\lambda _{2}$ then kinetic energy of electrons changes to $E_{2}$ . Then work function of the metal is :-

• A. $\frac{E_{1} E_{2}\left(\lambda_{1}-\lambda_{2}\right)}{\lambda_{1} \lambda_{2}}$
• B. $\frac{E_{1} \lambda_{2}-E_{2} \lambda_{2}}{\left(\lambda_{1} \lambda_{2}\right)}$
• C. $\frac{\left(E_{1} \lambda_{1}-E_{2} \lambda_{2}\right)}{\left(\lambda_{2}-\lambda_{1}\right)}$
• D. $\frac{\left(\lambda_{1} \lambda_{2} E_{1} E_{2}\right)}{\left(\lambda_{2}-\lambda_{1}\right)}$

Answer 1

Answer: (C)

For $\lambda_{1} \Rightarrow \frac{h c}{\lambda_{1}}-\phi_{1}=E_{1} \ldots \ldots .$ (i)
For $\lambda_{1} \Rightarrow \frac{h c}{\lambda_{2}}-\phi_{1}=E_{2}$
From (i) and (ii) $E_{1} \lambda_{1}+\phi_{1} \lambda_{1}=E_{2} \lambda_{2}+\phi_{1} \lambda_{2}$ $\Rightarrow \phi_{1}=\frac{E_{1} \lambda_{1}-E_{2} \lambda_{2}}{\left(\lambda_{2}-\lambda_{1}\right)}$