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Question: A photon is incident having frequency \(1\times {{10}^{14}}{{\sec }^{-1}}\) threshold frequency of m...

A photon is incident having frequency 1×1014sec⁡−11\times {{10}^{14}}{{\sec }^{-1}} threshold frequency of metal is 5×1013sec⁡−15\times {{10}^{13}}{{\sec }^{-1}}. Find the kinetic energy of the ejected electron?
A.3.3×10−21J3.3\times {{10}^{-21}}J
B.6.6×10−21J6.6\times {{10}^{-21}}J
C.3.3×10−20J3.3\times {{10}^{-20}}J
D.6.6×10−20J6.6\times {{10}^{-20}}J

Explanation

Solution

As a first step recall Einstein’s photoelectric equation based on energy quantum of radiation. Also, remember that threshold frequency is the minimum frequency required by the incident radiation to cause photoelectric emission. Now you could directly substitute the given values and hence get the answer.
Formula used:
Einstein’s photoelectric equation,
K.Emax⁡=hν−ϕ0K.{{E}_{\max }}=h\nu -{{\phi }_{0}}

Complete answer:
In the question we are given the frequency of the incident photon and the threshold frequency of metal is also given. With this given information we are asked to find the ejected electron.
We have Einstein’s photoelectric equation based on the energy quantum of radiation. The theory behind this equation is that, here the radiation energy is said to be composed of discrete units called quanta of energy of radiation. Each ‘quanta’ is known to have energy hνh\nu where h is the Planck’s constant and ν\nu is the frequency of light. During a photoelectric effect an electron absorbs this quantum of energy. Incase if the absorbed energy is greater than the work functionϕ0{{\phi }_{0}}, that is, the minimum energy required for the electron to escape from the surface of the metal, then the electron gets emitted with maximum kinetic energy.
Einstein’s photoelectric equation is mathematically expressed as,
K.Emax⁡=hν−ϕ0K.{{E}_{\max }}=h\nu -{{\phi }_{0}} ………………………………. (1)
Also, the work function ϕ0{{\phi }_{0}} can be expressed as,
ϕ0=hν0{{\phi }_{0}}=h{{\nu }_{0}}
Where, ‘ν0{{\nu }_{0}}’ is the threshold frequency or the minimum frequency required for photoelectric emission. Equation (1) can rewritten as,
⇒K.Emax⁡=h(ν−ν0)\Rightarrow K.{{E}_{\max }}=h\left( \nu -{{\nu }_{0}} \right) …………………… (2)
Given values in the question are:
ν=1×1014sec⁡−1\nu =1\times {{10}^{14}}{{\sec }^{-1}}
ν0=5×1013sec⁡−1{{\nu }_{0}}=5\times {{10}^{13}}{{\sec }^{-1}}
Also we know that,
h=6.626×10−34Jsh=6.626\times {{10}^{-34}}Js
Substituting these in (2) gives,
⇒K.Emax⁡=6.626×10−34(1×1014−5×1013)\Rightarrow K.{{E}_{\max }}=6.626\times {{10}^{-34}}\left( 1\times {{10}^{14}}-5\times {{10}^{13}} \right)
⇒K.Emax⁡=6.626×10−34×1014(1−0.5)\Rightarrow K.{{E}_{\max }}=6.626\times {{10}^{-34}}\times {{10}^{14}}\left( 1-0.5 \right)
⇒K.Emax⁡=6.626×10−20×0.5\Rightarrow K.{{E}_{\max }}=6.626\times {{10}^{-20}}\times 0.5
⇒K.Emax⁡=3.313×10−20J\Rightarrow K.{{E}_{\max }}=3.313\times {{10}^{-20}}J
Therefore, the kinetic energy of the ejected electron is 3.3×10−20J3.3\times {{10}^{-20}}J.

So, the correct answer is “Option C”.

Note:
The maximum kinetic energy of the emitted photoelectrons though varies linearly with the frequency of incident radiation doesn’t depend on its intensity. If the frequency of incident radiation is lower than threshold frequency, no photoelectric emission takes place even for very high intensity. This threshold frequency is however different for different metals.