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The work function ($\phi $) of some metals are listed below. The number of metals which will show the photoelectric effect when the light of $300$ nm wavelength falls on the metal is:
MetalLiNaKMgCuAgFePtW
($\phi $) eV$2.4$$2.3$$2.2$$3.7$$4.8$$4.3$$4.7$$6.3$$4.75$


A) $1$
B) $2$
C) $3$
D) $4$

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Last updated date: 13th Jun 2024
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Answer
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Hint: You should know that; for the occurrence of the photoelectric effect, the energy associated with the photon should be greater than that of the work function of the metal. So, first we will calculate the energy associated with $300$ nm and then see whose work function of the given metals are lesser than that value.

Complete step by step solution:
Before proceeding for the calculation, let us know about the photoelectric effect.
So, the photoelectric effect is a phenomenon where electrons from the surface of a metal are emitted by absorbing light. When a light with sufficient energy is incident on a metal surface, the electrons on the surface of the metals absorb this light and are emitted out of the surface of the metal.
- Thus, we can say that when light is incident on a metal surface, the photons collide with the free electrons and in this particular collision, the photon may emit all of its energy to the free electron. Therefore, a minimum amount of energy i.e. equal to the work function ($\phi $) is to be given to an electron so as to bring out the effect. If the incident energy is more than the work function ($\phi $) the electron will come out of the metal to show the photoelectric effect.
- For a given wavelength, the energy can be found by the below formula:
$E=\dfrac{hc}{\lambda }$, where
E is the energy,
h is the Planck’s constant,
c is the speed of the light and
$\lambda $ is the wavelength.

Now, let us proceed to the calculation part.
Given that,
The incident wavelength i.e. $\lambda $ is given as $300$ nm.
So, the incident energy in electron volts will be:
$E=\dfrac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{300\times {{10}^{-9}}\times 1.6\times {{10}^{-19}}}eV=4.14eV$

- We have found the incident energy as $4.14eV$. From the above notion, we concluded that for photoelectric effect to occur, the energy of a photon must be greater than the work function of the metal.
- Already the work functions of the metals are given in the given data. So, let us see which metals are having lesser value than the incident energy.
- We found that: Li, Na, K and Mg are having lower values than the incident energy value. So, these metals will show photoelectric effect. Therefore, four metals are showing photoelectric effect.
So, the correct answer is “Option D”.

Note: The opposite of photoelectric effect is also possible. When we have a beam of fast-moving electrons near any positively charged metal surface, it can absorb the electron emitting light of certain wavelength.