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Question: Let us assume that electrons are getting ejected from the surface of a metal, if the light of partic...

Let us assume that electrons are getting ejected from the surface of a metal, if the light of particular frequency is incident on it and they are stopped completely by a retarding potential of 3V3V. The photoelectric effect in this metallic surface will be started at a frequency 6×1014s−16\times {{10}^{14}}{{s}^{-1}}. What will be the frequency of the incident light in s−1{{s}^{-1}}? (h=6×10−34Js,charge of the electron=1.6×10−19C)\left( h=6\times {{10}^{-34}}Js,\text{charge of the electron}=1.6\times {{10}^{-19}}C \right)

& A.7.5\times {{10}^{13}} \\\ & B.13.5\times {{10}^{3}} \\\ & C.14\times {{10}^{14}} \\\ & D.7.5\times {{10}^{15}} \\\ \end{aligned}$$
Explanation

Solution

The stopping potential can be found by ratio of the difference of the energy of the light incidents and the threshold energy of the surface to the charge of the electron. The energy can be found by taking the product of the Planck’s constant and the frequency. Substitute the values in it and then rearrange the equation in terms of the frequency of the light. This will help you in answering this question.

Complete step by step answer:
the stopping potential can be found by ratio of the difference of the energy of the light incidents and the threshold energy of the surface to the charge of the electron. This can be written as,
V=(hν−hν0)eV=\dfrac{\left( h\nu -h{{\nu }_{0}} \right)}{e}
This equation can be written as,
V=h(ν−ν0)eV=\dfrac{h\left( \nu -{{\nu }_{0}} \right)}{e}
It has been mentioned in the question that,
The Planck’s constant be,
h=6×10−34Jsh=6\times {{10}^{-34}}Js
The charge of the electron can be written as,
e=1.6×10−19Ce=1.6\times {{10}^{-19}}C
Stopping potential has been given as,
V=3VV=3V
Substituting this values in the equation can be written as,
3×1.6×10−19=5×10−34(ν−ν0)3\times 1.6\times {{10}^{-19}}=5\times {{10}^{-34}}\left( \nu -{{\nu }_{0}} \right)
As the threshold frequency has been given as,
ν0=6×1014s−1{{\nu }_{0}}=6\times {{10}^{14}}{{s}^{-1}}
Substituting this in the equation can be written as,
3×1.6×10−195×10−34=(ν−6×1014)\dfrac{3\times 1.6\times {{10}^{-19}}}{5\times {{10}^{-34}}}=\left( \nu -6\times {{10}^{14}} \right)
This equation can be written as,
3×1.6×10−195×10−34+6×1014=v\dfrac{3\times 1.6\times {{10}^{-19}}}{5\times {{10}^{-34}}}+6\times {{10}^{14}}=v
The frequency can be written as,
v=14×1014s−1v=14\times {{10}^{14}}{{s}^{-1}}

So, the correct answer is “Option C”.

Note: The photoelectric effect is defined as the phenomenon of ejection of electrons if an electromagnetic radiation, such as light, incidents on a material. The electrons which are emitted in this way are known as photoelectrons. The threshold energy is the minimum energy required to make the ejection possible.