Question

The work function of a metal in $4eV$. For the emission of photoelectrons of zero velocity from the metal surface, the wavelength of the incident radiation should be

• A. $1700\stackrel{\circ }{A}$
• B. $2700\stackrel{\circ }{A}$
• C. $3100\stackrel{\circ }{A}$
• D. $5900\stackrel{\circ }{A}$

Answer 1

The correct option is C $3100\stackrel{\circ }{A}$

We know,

$KE=hv-\varphi$

$1/2{mv}^2=\frac{hc}{\lambda }-\varphi$

Velocity $=0$

$\frac{12400}{\lambda \left(A^0\right)}=4eV$

$\therefore$ $\lambda =\frac{12400}{4}=3100A^0$