Question

The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength $2000\stackrel{\circ }{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is
$[h=4.14\times {10}^{-15}eVsec]$

• A. 1.2 volts
• B. 2.24 volts
• C. 3.6 volts
• D. 4.8 volts

Answer 1

The correct option is A 1.2 volts

Energy of incident light $E=\frac{12375}{2000}=6.18eV$
According to relation $E=W_0+e{V}_0$
$\Rightarrow V_0=\frac{(E-W_0)}{e}=\frac{(6.18eV-5.01eV)}{e}=1.17V\approx 1.2V$