Q. For a certain metal $v = 2\ v_{0}$ and the electrons come out with a maximum velocity of $4 \times 10^{6}\ ms^{- 1}$. If the value of $v = 5\ v_{0}$, then maximum velocity of photoelectrons will be
A) $2 \times 10^{7}\ ms^{- 1}$
B) $8 \times 10^{6}\ ms^{- 1}$
C) $2 \times 10^{6}\ ms^{- 1}$
D) $8 \times 10^{5}\ ms^{- 1}$
Explanation
(b) $\frac{1}{2}mv_{1}^{2} = 2hv_{0} - hv_{0} = hv_{0}$ and
$\frac{1}{2}mv_{2}^{2} = 5hv_{0} - hv_{0} = 4\ hv_{0}$
So, $\frac{1}{2}mv_{2}^{2} = 4 \times \frac{1}{2}mv_{1}^{2}$
or $v_{2} = 2v_{1} = 2 \times 4 \times 10^{6} = 8 \times 10^{6}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.