Question

For a certain metal, incident frequency $v$ is five times of threshold frequency $v_0$ and the maximum velocity of coming out photoelectrons is $8\times 10^{6}m{s}^{-1}$. If $v=2v_0$, 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)

According to Einstein’s photoelectric equation
$h\upsilon =h{\upsilon }_0+\frac{1}{2}m{\upsilon }_{max}^2$
or $\frac{1}{2}m{\upsilon }_{max}^2=h\upsilon -h{\upsilon }_0$
According to the given problem
$\frac{1}{2}m(8\times 10^6{)}^2=h(5{\upsilon }_0-={\upsilon }_0)$ ....(i)
$\frac{1}{2}m{\upsilon }_{max}^2=h(2{\upsilon }_0-h{\upsilon }_0)$ ....(ii)
Dividing equation (i) by (ii)
$\frac{(8\times 10^6{)}^2}{{\upsilon }_{max}^2}=\frac{4{\upsilon }_0}{{\upsilon }_0}$
${\upsilon }_{max}^2=\frac{(8\times 10^6{)}^2}{4}$
${\upsilon }_{max}=\frac{8\times 10^6}{2}$
= $4\times 10^{6}m{s}^{-1}$