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# 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:(A) 1:2(B) 1:1(C) 1:5(D) 1:4

Last updated date: 12th Jul 2024
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Hint: The maximum kinetic energy of the electrons is equal to the energy of the radiations reduced by the work function (i.e. energy of photon minus work function of metal). Kinetic energy is proportional to the square of the speeds,
Formula used: In this solution we will be using the following formulae;
$K{E_{\max }} = E - W$ where $K{E_{\max }}$ is the maximum kinetic energy of the ejected electrons, $E$ is the energy of the photons of the radiation, and $W$ is the work function of the metal.

Complete Step-by-Step Solution:
Two different radiations are said to illuminate a metallic surface of a particular work function, we are to determine the ratio of the kinetic energy of the electrons ejected from the metal.
To do so, we must at first calculate the kinetic energy of the photons in the individual cases.
The formula for the kinetic energy is given by
$K{E_{\max }} = E - W$ where $K{E_{\max }}$where $E$ is the energy of the photons of the radiation, and $W$ is the work function of the metal.
Hence, for the first radiation, we have
$K{E_{\max 1}} = 1eV - 0.5eV$
$\Rightarrow K{E_{\max 1}} = 0.5eV$
For the second radiation, we have,
$K{E_{\max 2}} = 2.5eV - 0.5eV$
$\Rightarrow K{E_{\max 2}} = 2eV$
Hence, the ratio will be given as
$\dfrac{{K{E_{\max 1}}}}{{K{E_{\max 2}}}} = \dfrac{{0.5}}{2} = \dfrac{1}{4}$
But Kinetic energy is proportional to the square of the speeds, then
$\dfrac{{{v_1}^2}}{{{v_2}^2}} = \dfrac{1}{4}$
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{1}{4}} = \dfrac{1}{2}$
Hence, the ratio of one to the other is
${v_1}:{v_2} = 1:2$

Thus, the correct option is A

Note: We need to observe that to find the ratio of the two kinetic energies, the unit does not have to be converted to SI to get the proper answer. This is because the conversion factor will end up cancelling out, and the values only will matter. Similarly, for replacing kinetic energy with just the square of the speeds, the constants will cancel out anyway.