Question

A light source of wavelenght $\lambda$ illuminates a metal and ejects photo-electrons with $(K.E.{)}_{max}=1.eV$. Another light source of wavelength $\frac{\lambda }{3}$, ejects photo-electrons from same metal with $(K.E.{)}_{max}=4eV$. Find the value of work function.

• A. $1eV$
• B. $2eV$
• C. $0.5eV$
• D. $1.5eV$

Answer 1

The correct option is C $0.5eV$

Let work function be $\varphi$

Then ; ${(K.E)}_{max}=\frac{hc}{\lambda }-\varphi \therefore 4ev=\frac{hc}{\lambda }-\varphi -\left(i\right)4ev=\frac{3hc}{\lambda }-\varphi -\left(ii\right)$

$\therefore$ Replacing $\frac{hc}{\lambda }$ in $\left(ii\right)$ by $\left(i\right)$;

$4ev=(1+\varphi )3-\varphi \Rightarrow 2\varphi =1ev\therefore \varphi =0.5ev$