Question

A photosensitive metallic surface has work function $ \phi $. If photon of energy 3 $ \phi $ fall on this surface, the electron comes out with a maximum velocity of $ 6 \times 10^6 \, ms^{ - 1} $.
When the photon energy is increased to 9 $ \phi $, there maximum velocity of photoelectron will be

• A. $ 12 \times 10^6 \, m/s $
• B. $ 6 \times 10^6 \, m/s $
• C. $ 3 \times 10^6 \, m/s $
• D. $ 24 \times 10^6 \, m/s $

Answer 1

Answer: (A)

The relation between the energy and the work function of the electron is given as:
$E =\phi+\frac{1}{2} m v ^{2}$
For the total energy of $3 \phi$, the expression can be written as:
$3 \phi=\phi+\frac{1}{2} m v _{1}^{2}$
$\frac{1}{4} mv _{1}^{2}=\phi$
Now, the energy provided is $9 \phi .$ So, it can be written as:
$9 \phi=\phi+\frac{1}{2} m v _{2}^{2}$
$8 \phi=\frac{1}{2} mv _{2}^{2}$
Substitute the value of $\phi$.
$8 \times \frac{1}{4} m v _{1}^{2}=\frac{1}{2} m v _{2}^{2}$
$v _{2}=\sqrt{ 4 v _{ 1 }^{2}}$
$\Rightarrow \sqrt{4 \times 36 \times 10^{12}}$
$12 \times 10^{6} m / s$