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 $v=2 v_0$, 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
$h \upsilon = h \upsilon_0 + \frac{1}{2} m\upsilon^2_{max}$
or $ \frac{1}{2} m\upsilon^2_{max} = 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^2_{max} = h(2 \upsilon_0 - h \upsilon_0)$ ....(ii)
Dividing equation (i) by (ii)
$ \frac{(8 \times 10^6)^2}{\upsilon^2_{max}} = \frac{4 \upsilon_0}{\upsilon_0}$
$\upsilon^2_{max} = \frac{(8 \times 10^6)^2}{4}$
$ \upsilon_{max} = \frac{8 \times 10^6}{2}$
= $4 \times 10^6 \, m \, s^{-1}$