Question

For a certain metal, incident frequency $\upsilon$ is five times of threshold frequency $\upsilon$ and the maximum velocity of coming out photoelectrons is $8\times {10}^{6}m{s}^{-1}$ If $\upsilon =2{\upsilon }_o$, the maximum velocity of photoelectrons will be

• A. $4\times {10}^{6}m{s}^{-1}$
• B. $6\times {10}^{6}m{s}^{-1}$
• C. $8\times {10}^{6}m{s}^{-1}$
• D. $1\times {10}^{6}m{s}^{-1}$

Answer 1

Answer: (A)

$4\times {10}^{6}m{s}^{-1}$

According to Einsteins photoelectrical equation
$h\upsilon =h{\upsilon }_0+\frac{1}{2}m{v}_{max}^{2}or\frac{1}{2}m{v}_{max}^2=h\upsilon -h{\upsilon }_0$

According to the given problem
$\frac{1}{2}m(8\times {10}^6{)}^2=h(5{\upsilon }_0-{\upsilon }_0)=4h{\upsilon }_0$ ................(i)

$\frac{1}{2}m{v}_{max}^2=h(2{\upsilon }_0-{\upsilon }_0)=h{\upsilon }_0$ .....................(ii)

Dividing Equation (i) by (ii) we get

$\frac{(8\times {10}^6{)}^2}{v_{max}^2}=\frac{4{\upsilon }_0}{{\upsilon }_0}\Rightarrow v_{max}^2=\frac{(8\times {10}^6{)}^2}{4}$

$v_{max}=\frac{8\times {10}^6}{2}=4\times {10}^{6}m{s}^{-1}$