Question

A radiation of wavelength $\lambda$ illuminates a metal and ejects photoelectrons of maximum kinetic energy of $1eV$. Another radiation of wavelength $\frac{\lambda }{3}$, ejects photoelectrons of maximum kinetic energy of $4eV$. What will be the work function of metal?

• A. $1eV$
• B. $2eV$
• C. $0.5eV$
• D. $3eV$

Answer 1

The correct option is C $0.5eV$

${KE}_{max}$ =$\frac{hc}{\lambda }$ - $\varphi$

$\varphi$ is work function

$\lambda$ is wavelength

h is Plank's constant

c is speed of light

Make two equations according to given data :

1 =$\frac{hc}{\lambda }$ - $\varphi$ ............(eqn 1)

4=$\frac{hc}{\frac{\lambda }{3}}$ - $\varphi$ ........(eq2)

Solve both to get value of $\varphi$ we get 0.5 eV