Question

When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8\, V$. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6\, V$. 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}=h v-\phi $
$4.8=\frac{h c}{\lambda}-\phi$ ...(i)
$1.6=\frac{h c}{2 \lambda}-\phi$ ... (ii)
Using above equation (i) - (ii)
$3.2=\frac{h c}{\lambda}-\frac{h c}{2 \lambda}$
$3.2=\frac{h c}{2 \lambda}$ ... (iii)
$\left[\lambda=\frac{h c}{6.4}\right]$
Put in equation (ii)
$\phi=1.6$
$\frac{h c}{\lambda_{t h}}=1.6$
$\lambda_{t h}=\frac{h c}{1.6}$
$=\left(\frac{h c}{6.4}\right) \times 4=4 \lambda$