Question

The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from $400$ nm to $310$ nm. The work function of the metal is :

• A. $0.9eV$
• B. $1.7eV$
• C. $2.2eV$
• D. $3.1eV$

Answer 1

The correct option is C $2.2eV$

The Einstein's equation for photoelectric effect : $K=\frac{hc}{\lambda }-W$

Now, $K=\frac{hc}{{\lambda }_1}-W....\left(1\right)$

and $2K=\frac{hc}{{\lambda }_2}-W....\left(2\right)$

So, $\frac{2hc}{{\lambda }_1}-2W=\frac{hc}{{\lambda }_2}-W$ [using $\left(1\right)$]

or $W=hc[\frac{2}{{\lambda }_1}-\frac{1}{{\lambda }_2}]=(6.62\times {10}^{-34})(3\times {10}^8)[2/400-1/310]\times {10}^9=3.52\times {10}^{-19}J=2.2eV$ as

$[1eV=1.6\times {10}^{-19}J]$