Question

Photoelectric work function of a metal 1 eV. If a light of wavelength 500 Å falls on the metal, the speed of the photoelectron emitted approximately will be- (h = 6.625 × ${10}^{–34}$ J s)

• A. Equal to velocity of light
  
  
  
• B. $\frac{1}{10}\times$ velocity of light
  
• C. $\frac{1}{100}\times$ velocity of light
  
• D. $\frac{1}{1000}\times$ velocity of light

Answer 1

The correct option is C

$\frac{1}{100}\times$ velocity of light

$hv=w+KE=w+\frac{1}{2}{m}_e{v}^{2}orh\frac{c}{\lambda }=1+\frac{1}{2}{m}_e{v}^{2}or\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-10}}=1.6\times {10}^{-19}+\frac{1}{2}\times 9.1\times {10}^{-31}\times v^{2}or\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=\frac{6.625\times 3}{5}\times {10}^{-18}-1.6\times {10}^{-19}=38.15\times {10}^{-19}or{v}^2=\frac{38.15\times 2\times {10}^{12}}{9.1}or{v}^2=8.385\times {10}^{12}orv=2.889\times {10}^{6}m/s\therefore \frac{v}{c}\simeq \frac{3\times {10}^6}{3\times {10}^8}=\frac{1}{100}$