Question

A metal surface is illuminated by light of two different wavelengths $248nm$ and $310nm.$ The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_{2,}$ respectively. If the ratio$u_1:u_2=2:1$ and $hc=1240eVnm,$ the work function of the metal is nearly

• A. $2.5eV$
• B. $3.7eV$
• C. $3,2eV$
• D. $2.8eV$

Answer 1

The correct option is B $3.7eV$

$\frac{1}{2}m{u}_1^2=\frac{hc}{{\lambda }_1}-W$

$\frac{1}{2}m{u}_2^2=\frac{hc}{{\lambda }_2}-W$

${\left(\frac{u_1}{u_2}\right)}^2=\frac{\frac{hc}{{\lambda }_1}-W}{\frac{hc}{{\lambda }_2}-W}$

$\frac{4hc}{{\lambda }_2}-4W=\frac{hc}{{\lambda }_1}-W(\because \frac{u_1}{u_2}=2)$

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

$\Rightarrow \frac{4\times 1240}{310}-\frac{1240}{248}=3W$
$\Rightarrow 16-5=3W$
$\Rightarrow 11=3W$
$\Rightarrow W=3.7eV$