Question

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

• A. 1.2 V
• B. 2.4 V
• C. 3.6 V
• D. 4.8 V

Answer 1

The correct option is A 1.2 V

The Einstein's equation for photoelectric effect is given by: $K_{max}=h\nu -W$

$e{V}_0=\frac{hc}{\lambda }-W$

$V_0=\frac{(hc/\lambda -W)}{e}=\frac{\left[\right(6.62\times {10}^{-34}\times 3\times {10}^8\left)/\right(2000\times {10}^{-10})-(5.01\times 1.6\times {10}^{-19}\left)\right]}{1.6\times {10}^{-19}}$ $=1.19V=1.2V$