Question

If the wavelength ${\lambda }_1$ is allowed to fall on metal, then kinetic energy of photoelectrons emitted is $E_1$. If wavelength of light changes to ${\lambda }_2$ then kinetic energy of electrons also changes to $E_2$. Then work function of the metal is

• A. $\frac{E_1{E}_2({\lambda }_1-{\lambda }_2)}{{\lambda }_1{\lambda }_2}$
• B. $\frac{E_1{\lambda }_1-E_2{\lambda }_2}{{\lambda }_2-{\lambda }_1}$
• C. $\frac{E_1{\lambda }_1-E_2{\lambda }_2}{({\lambda }_1-{\lambda }_2)}$
• D. $\frac{{\lambda }_1{\lambda }_2{E}_1{E}_2}{{\lambda }_2-{\lambda }_1}$

Answer 1

Answer: (B)

$\frac{E_1{\lambda }_1-E_2{\lambda }_2}{{\lambda }_2-{\lambda }_1}$

As we know,

$E=W_0+K_{max}\Rightarrow \frac{hc}{{\lambda }_1}=W_0+E_1$ and

$\frac{hc}{{\lambda }_2}=W_0+E_2$

$\Rightarrow hc=W_0{\lambda }_1+E_1{\lambda }_1$ and $hc=W_0{\lambda }_2+E_2{\lambda }_2$

$\Rightarrow W_0{\lambda }_1+E_1{\lambda }_1=W_0{\lambda }_2+E_2{\lambda }_2\Rightarrow W_0=\frac{E_1{\lambda }_1-E_2{\lambda }_2}{({\lambda }_2-{\lambda }_1)}.$