Question

When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8V$. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6V$. The threshold wavelength of the metal is :

• A. $2\lambda$
• B. $4\lambda$
• C. $8\lambda$
• D. $6\lambda$

Answer 1

Answer: (B)

$V_s=hv-\varphi$
$4.8=\frac{hc}{\lambda }-\varphi$ ...(i)
$1.6=\frac{hc}{2\lambda }-\varphi$ ... (ii)
Using above equation (i) - (ii)
$3.2=\frac{hc}{\lambda }-\frac{hc}{2\lambda }$
$3.2=\frac{hc}{2\lambda }$ ... (iii)
$\left[\lambda =\frac{hc}{6.4}\right]$
Put in equation (ii)
$\varphi =1.6$
$\frac{hc}{{\lambda }_{th}}=1.6$
${\lambda }_{th}=\frac{hc}{1.6}$
$=\left(\frac{hc}{6.4}\right)\times 4=4\lambda$