Question

The surface of certain metal is first illuminated with light of wavelength ${\lambda }_1=350\text{nm}$ and then, by light of wavelength ${\lambda }_2=540\text{nm}$. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of $2$. The work function of the metal $\text{(in eV)}$ is close to:
$\text{(Energy of photon}=\frac{1240}{\lambda \left(\text{in nm}\right)}\text{eV})$

• A. $5.6$
• B. $1.8$
• C. $2.5$
• D. $1.4$

Answer 1

The correct option is B $1.8$

From Einstein's photoelectric equation,

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi$

$\text{Given that,}{v}_1=2v_2$

$\text{For case -1}$
$\frac{1}{2}m(2v_2{)}^2=\frac{hc}{{\lambda }_1}-\varphi ...(1)$

$\text{For case -2}$
$\frac{1}{2}m(v_2{)}^2=\frac{hc}{{\lambda }_2}-\varphi ...(2)$

Dividing $\left(1\right)$ by $\left(2\right)$

$\Rightarrow \frac{\frac{hc}{{\lambda }_1}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }=4$

$\Rightarrow \frac{hc}{{\lambda }_1}-\varphi =\frac{4hc}{{\lambda }_2}-4\varphi$

$\Rightarrow 3\varphi =\frac{4hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}$

$\Rightarrow \varphi =\frac{1}{3}(4\times \frac{1240\text{eV nm}}{540\text{nm}}-\frac{1240\text{eV nm}}{350\text{nm}})$

$\Rightarrow \varphi =\frac{1}{3}(9.18-3.54)$

$\Rightarrow \varphi =1.8\text{eV}$

Hence, option $\left(A\right)$ is correct.