Question

Consider a metal exposed to light of wavelength $600nm$. The maximum energy of the electron doubles when light of wavelength $400nm$ is used. Find the work function in $eV$.

Answer 1

Formula Used:
$W_0=h{v}_0=\frac{hc}{{\lambda }_0}Joules$;
$v_0=\text{Thersold frequency}$
Maximum energy$=hv-\varphi$
Work Function in electron volt,
$W_0\left(eV\right)=\frac{hc}{e{\lambda }_0}=\frac{12400}{{\lambda }_0\left(A^{\circ }\right)}$

Given, ${\lambda }_1=600,{\lambda }_2=400nm$

Maximum kinetic energy for the second condition is equal to the twicw of the kinetic energy in first condition.

i.e., $K_{max}^{{}^{\prime }}=2K_{max}$

then $K_{max}^{{}^{\prime }}=\frac{hc}{\lambda }-\varphi$
$\Rightarrow 2K_{max}=\frac{hc}{{\lambda }^{{}^{\prime }}}-{\varphi }_0$
$\Rightarrow 2(\frac{1240}{600}-\varphi )=(\frac{1240}{400}-\varphi )$
$[\because hc=1240eVnm]$
$\varphi =1240[\frac{1}{300}-\frac{1}{400}]$
$\Rightarrow \varphi =1.02eV$

Final Answer: $1.02eV$