Question

A photosensitive metallic surface has work function $\varphi$. If photon of energy $3\varphi$ fall on this surface, the electron comes out with a maximum velocity of $6\times {10}^{6}m/s$. When the photon energy is increased to $9\varphi$, then 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

The correct option is A $12\times {10}^{6}m/s$

The relation between the energy and the work function of the electron is given as:

$E=\varphi +\frac{1}{2}m{v}^2$

For the total energy of $3\varphi$, the expression can be written as:

$3\varphi =\varphi +\frac{1}{2}m{v}_1^2$

$\frac{1}{4}m{v}_1^2=\varphi$

Now, the energy provided is $9\varphi$. So, it can be written as:

$9\varphi =\varphi +\frac{1}{2}m{v}_2^2$

$8\varphi =\frac{1}{2}m{v}_2^2$

Substitute the value of $\varphi$.

$8\times \frac{1}{4}m{v}_1^2=\frac{1}{2}m{v}_2^2$

$v_2=\sqrt{4v_1^2}$

$\Rightarrow \sqrt{4\times 36\times {10}^{12}}$

$12\times {10}^{6}m/s$