Question

Two radiations of photons energies 1eV and 2.5eV, successively illuminate a photosensitive metallic surface of work function 0.5eV . The ratio of the maximum speeds of the emitted electrons is:
1. 1:2
2. 1:1
3. 1:5
4. 1:4

Answer 1

Energy of photon is given by:
$E = h\upsilon$ (h is planck's constant , $\upsilon$ is the frequency of photon)
Using Einstein energy equation:
Energy of photon = kinetic energy-work function.
${K_e} = E - \omega$ (E, is the energy of photon, Ke is the kinetic energy and w is the work function)
${K_e} = \dfrac{1}{2}m{v^2}$
Using the above relation we will solve the problem.

Complete step by step solution:
First we will define the work function and photon.
Work function: the minimum amount of energy required to eject an electron out of the metal surface is called the work function of the metal.
Photon: light consists of particles associated with a definite amount of energy and momentum .These particles were named as Photon.

Now, come to the mathematical part of the problem :
$v_1$ and $v_2$ are speeds of two different radiations.
$\Rightarrow \dfrac{1}{2}m{v_1}^2 = hv - w \\\ \Rightarrow \dfrac{1}{2}m{v_2}^2 = hv - w \\\$ (using the relation we have discussed in the hint )

Putting the numerical values of the formula
$\Rightarrow \dfrac{1}{2}m{v_1}^2 = 1 - 0.5 = 0.5 \\\ \Rightarrow \dfrac{1}{2}m{v_2}^2 = 2.5 - 0.5 = 2 \\\$

Now dividing both the equations:
$\Rightarrow \dfrac{{\dfrac{1}{2}m{v_1}^2}}{{\dfrac{1}{2}m{v_2}^2}} = \dfrac{{0.5}}{2} \\\ \Rightarrow \dfrac{{{v_1}^2}}{{{v_2}^2}} = \dfrac{1}{4} \\\$( cancelling m and ½ )
$\Rightarrow \dfrac{{{v_{_1}}}}{{{v_2}}} = \sqrt {\frac{1}{4}} \\\ \Rightarrow \dfrac{{{v_{_1}}}}{{{v_2}}} = \dfrac{1}{2} \\\$

Thus , option 1 is correct.

Note:
Kinetic Energy of the photon does not depend on the intensity of the light wave, rather it depends on the frequency of the light wave. Photoelectrons emitted per second do depend on the intensity of the light wave and not on the frequency of the light. The frequency of light must be greater than the threshold frequency $f_0$.