Question

A photon of energy 2.5 eV and wavelength ${}^{\prime }{\lambda }^{\prime }$ falls on a metal surface and the ejected electrons have maximum velocity ${}^{\prime }{v}^{\prime }$. If the ${}^{\prime }{\lambda }^{\prime }$ of the incident light is decreased by 20%, the maximum velocity of the emitted electrons is doubled. The work function of the metal is :

• A. 2.6 eV
• B. 2.23 eV
• C. 2.5 eV
• D. 2.29 eV

Answer 1

The correct option is D 2.29 eV

Energy $=\frac{hc}{\lambda }=2.5ev$

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

$\frac{1}{2}m(2v{)}^2=\frac{hc5}{4\lambda }-\varphi$ -------------(2)

from 2
$4\left(\frac{1}{2}m{v}^2\right)=\frac{5hc}{4\lambda }-\varphi$

Now, using 1
$4(\frac{hc}{\lambda }-\varphi )=\frac{5hc}{4\lambda }-\varphi$

$\frac{4hc}{\lambda }-4\varphi =\frac{5hc}{4\lambda }-\varphi$

$\frac{11hc}{4\lambda }=3\varphi$

$\frac{11hc}{12\lambda }=\varphi$

$\varphi =\frac{11}{12}\left(2.5\right)$
$=2.29eV.$

So, the answer is (D).