Q94 / 160 · JIPMER 2007 · Physics
A photosensitive metallic surface has work function, $h v_0$. If photons of energy $2 h v_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 \mathrm{~m} / \mathrm{s}$. When the photon energy is increased to $5 h v_0$, then maximum velocity of photoelectrons will be
- A$2 \times 10^6 \mathrm{~m} / \mathrm{s}$
- B$2 \times 10^7 \mathrm{~m} / \mathrm{s}$
- C$8 \times 10^5 \mathrm{~m} / \mathrm{s}$
- Correct answerD$8 \times 10^6 \mathrm{~m} / \mathrm{s}$
Solution
The solution to our problem consists in Einstein's photoelectric equation.
Einstein's photoelectric equation can be written as $\frac{1}{2} m v^2=h v-\phi$
$\Rightarrow \frac{1}{2} m \times\left(4 \times 10^6\right)^2=2 h v_0-h v_0...(i)$
and $\frac{1}{2} m \times v^2=5 h v_0-h v_0...(ii)$
Dividing Eq. (ii) by Eq. (i), we get
$\frac{v^2}{\left(4 \times 10^6\right)^2}=\frac{4 h v_0}{h v_0}$
$\begin{array}{ll}\Rightarrow & v^2=4 \times 16 \times 10^{12} \\ \Rightarrow & v^2=64 \times 10^{12} \\ \therefore & v=8 \times 10^6 \mathrm{~m} / \mathrm{s}\end{array}$
NOTE: The efficiency of photoelectric effect is less than $1 \%$.ie, number of photons less than $1 \%$ are capable of ejecting photoelectrons.
Hence, the correct answer is option: 'D'.