Question

If the maximum kinetic energy of emitted photo electrons from a metal surface of work function 2.5 eV, is 1.7 eV. If wavelength of incident radiation is halved, then stopping potential will be

• A. 2.5 V
• B. 6.7 V
• C. 5 V
• D. 1.1 V

Answer 1

Answer: (B)

Energy of incident photons,
$h\upsilon ={\varphi }_0+K_{max}=2.5+1.7=4.2eV$
$\because \upsilon \lambda =cand{\upsilon }^{{}^{\prime }}\frac{\lambda }{2}=c\Rightarrow {\upsilon }^{{}^{\prime }}=2\upsilon$
According to Einstein’s photoelectric equation,
$e{V}_0=h{\upsilon }^{{}^{\prime }}-{\varphi }_0=h\left(2\upsilon \right)-{\varphi }_0=2\times 4.2-1.7$
$=6.7eV\Rightarrow V_0=6.7V$