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 ${\lambda }_1\Rightarrow \frac{hc}{{\lambda }_1}-{\varphi }_1=E_1\dots \dots .$ (i)
For ${\lambda }_1\Rightarrow \frac{hc}{{\lambda }_2}-{\varphi }_1=E_2$
From (i) and (ii) $E_1{\lambda }_1+{\varphi }_1{\lambda }_1=E_2{\lambda }_2+{\varphi }_1{\lambda }_2$ $\Rightarrow {\varphi }_1=\frac{E_1{\lambda }_1-E_2{\lambda }_2}{\left({\lambda }_2-{\lambda }_1\right)}$