Question

The work function of cesium is $2.14\,eV$ The threshold frequency of caesium is, find the wavelength of the incident light if the stopping potential is $0.6\, V$

• A. $326\, nm$
• B. $454\, nm$
• C. $524\, nm$
• D. $232\, nm$

Answer 1

Answer: (B)

According to Einstein’s photoelectric equation
$eV_{0}=h\upsilon-\phi_{0}=\frac{hc}{\lambda}-\phi_{0} \left(\because\upsilon=\frac{c}{\lambda}\right)$
or $\lambda=\frac{hc}{\left(eV_{0}+\phi_{0}\right)}=\frac{1242\,eV\,nm}{\left(0.6+2.14\right)eV} \approx454\,nm$