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It takes $4.6eV$ to remove one of the least tightly bound electrons from a metal surface. When monochromatic photons strike the metal surface, electrons having kinetic energy from zero to $2.2eV$ are ejected. What is the energy of the incident photons?
A. $2.4eV$
B. $2.2eV$
C. $6.8eV$
D. $4.6eV$
E. $5.8eV$

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Last updated date: 01st Mar 2024
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Hint: When a photon falls on the surface of metal, the energy of the photon is transferred to the electron. A part of this energy will be acquired by the electrons from the metal and another part is acquired by the ejected electrons, which is known as kinetic energy. The formula used for calculating the energy of incident photons is given below.

Formula used:
The formula used for calculating the energy of the incident photons is given by
$E = W + K.E.$
Here, $E$ is the energy of the incident photons, $W$ is the work function of the metal surface and $K.E.$ is the kinetic energy acquired by the ejected electrons.

Complete step by step answer:
As given in the question, $4.6\,eV$ is required to remove least tightly bound electrons from a metal surface. Therefore, the work function of the metal surface is $W = 4.6eV$.
Also, the kinetic energy acquired by the ejected electrons from the metal surface is $2.2eV$. Therefore, the kinetic energy of ejected electrons is $K.E. = 2.2eV$
The formula used for calculating the energy of the incident photons is given by
$E = W + K.E.$
Here, $E$ is the energy of the incident photons, $W$ is the work function of the metal surface and $K.E.$ is the kinetic energy acquired by the ejected electrons.
$ \Rightarrow \,E = 4.6 + 2.2$
$ \therefore \,E = 6.8eV$
Therefore, the energy of the incident photons is $6.8eV$.

Hence, option C is the correct answer.

Note:Here, the formula used for calculating the energy of the incident photons is given by Einstein. Here, in the question, the values of work function and kinetic energy are given, that is why we have used this formula. Both the values are given in $eV$, therefore, we have not changed the units.
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