Question

When radiation is incident on a photo electron emitter, the stopping potential is found to be 9 V. If $\left(\frac{e}{m}\right)$ for the electron is $1.8\times {10}^{11}ck{g}^{-1}$, the maximum velocity of the ejected electron is

• A. $6\times {10}^{5}m{s}^{-1}$
  
• B. $8\times {10}^{5}m{s}^{-1}$
  
• C. $1.8\times {10}^{6}m{s}^{-1}$
  
• D. $0.8\times {10}^{3}m{s}^{-1}$
  

Answer 1

The correct option is C $1.8\times {10}^{6}m{s}^{-1}$


Here, $\frac{1}{2}m{V}_{max}^2=e{V}_0\Rightarrow V_{max}=\sqrt{\frac{2e{V}_0}{m}}=\sqrt{2\times 1.8\times {10}^{11}\times 9}=1.8\times {10}^{6}m{s}^{-1}$