Question

When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm, the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is

• A. $3.972\times {10}^{-19}J$
• B. $1.59\times {10}^{-19}J$
• C. $2\times {10}^{-19}J$
• D. $0.5\times {10}^{-19}J$

Answer 1

The correct option is A $3.972\times {10}^{-19}J$

$K.E.max=\frac{hc}{\lambda }-\varphi$

$\frac{1}{2}m{V}^2=\frac{hc}{\lambda }-\varphi$ $(\because K.E.max=\frac{1}{2}m{V}^2)$

$\frac{1}{2}m{V}^2=\frac{hc}{400}-\varphi$ -(1)

$\frac{1}{2}m(2V{)}^2=\frac{hc}{250}-\varphi$ -(2)

$4\left(\frac{1}{2}m{v}^2\right)=\frac{hc}{250}-\varphi$ -(3)

Now Rutting value of $\frac{1}{2}m{V}^2$ from (1) en to (3)

$4(\frac{hc}{400}-\varphi )=\frac{hc}{250}-\varphi$

$\frac{4hc}{400}-4\varphi =\frac{hc}{250}-\varphi$

$\frac{3}{2}\left(\frac{hc}{250}\right)=3\varphi$

$\varphi =\frac{1240}{2\left(250\right)}$

=2.48eV
=$2.48\times 1.6\times {10}^{-19}J$
=$3.972\times {10}^{-19}J$

So, the answer is option (A).