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 $6\times 10^{6}m{s}^{-1}$.
When the photon energy is increased to 9 $\varphi$, 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:
$E=\varphi +\frac{1}{2}m{v}^2$
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 $\varphi$.
$8\times \frac{1}{4}m{v}_1^2=\frac{1}{2}m{v}_2^2$
$v_2=\sqrt{4v_1^2}$
$\Rightarrow \sqrt{4\times 36\times 10^{12}}$
$12\times 10^{6}m/s$