Question

A photosensitive metallic surface has work function, $h{v}_0$. If photons of energy $2h{v}_0$ fall on this surface, the electrons come out with a maximum velocity of $4\times 10^{6}m/s$. When the photon energy is increased to $5h{v}_0$, then maximum velocity of photoelectrons will be

• A. $2\times 10^{6}m/s$
• B. $2\times 10^{7}m/s$
• C. $8\times 10^{5}m/s$
• D. $8\times 10^{6}m/s$

Answer 1

Answer: (D)

The solution to our problem consists, in Einsteins photoelectric equation.
Einsteins photoelectric equation can be written as
$\frac{1}{2}m{v}^2=hv-\varphi$
$\Rightarrow \frac{1}{2}m\times {\left(4\times 10^6\right)}^2=2h{v}_0-h{v}_0...(i)$
and $\frac{1}{2}m\times v^2=5h{v}_0-h{v}_0....(ii)$
Dividing Eq. (ii) by (i), we get
$\frac{v^2}{{\left(4\times 10^6\right)}^2}=\frac{4h{v}_0}{h{v}_0}$
$\Rightarrow v^2=4\times 16\times 10^{12}$
$\Rightarrow v^2=64\times 10^{12}$
$\therefore v=8\times 10^{6}m/s$
NOTE: The efficiency of photoelectric effect is less than $1\%$ i.e., number of photons less than $1\%$ are capable of ejecting photoelectrons.