Question

When a metal surface is illuminated by light of wavelengths $400\text{nm}$ and $250\text{nm},$ the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is$-[h=$Planck's constant, $c=$speed of light in vacuum$]$

• A. $2hc\times {10}^6\text{J}$
• B. $1.5hc\times {10}^6\text{J}$
• C. $hc\times {10}^6\text{J}$
• D. $0.5hc\times {10}^6\text{J}$

Answer 1

The correct option is A $2hc\times {10}^6\text{J}$

Given,
${\lambda }_1=400\text{nm};{\lambda }_2=250\text{nm}(v_1{)}_{max}=v;(v_2{)}_{max}=2v$

From Einstein's photoelectric equation,

$E=\varphi +(KE{)}_{max}$
$\frac{hc}{\lambda }=\varphi +\frac{1}{2}m{v}_{max}^2$
$\frac{1}{2}m{v}_{max}^2=\frac{hc}{\lambda }-\varphi$

For Case $I:$

$\frac{1}{2}m{v}^2=\frac{hc}{{\lambda }_1}-\varphi ......\left(1\right)$

For Case $II:$

$\frac{1}{2}m(2v{)}^2=\frac{hc}{{\lambda }_2}-\varphi .......(2)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\Rightarrow \left(\frac{1}{4}\right)=\frac{\frac{hc}{{\lambda }_1}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }$

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

$\varphi =\frac{hc}{3}[\frac{4}{{\lambda }_1}-\frac{1}{{\lambda }_2}]$

$=\frac{hc}{3}[\frac{4}{400\times {10}^{-9}}-\frac{1}{250\times {10}^{-9}}]$

$=\frac{hc}{3}\times {10}^9\times \frac{150}{100\times 250}$

$=2hc\times {10}^6\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.