
For photoelectric emission from certain metals the cut-off frequency is $v$. If the radiation of frequency $2v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be: (m is the mass of electron)
A. -\[\sqrt{\dfrac{hv}{2m}}\]
B. \[\sqrt{\dfrac{hv}{m}}\]
C. \[\sqrt{\dfrac{2hv}{m}}\]
D. \[2\sqrt{\dfrac{hv}{m}}\]
Answer
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Hint: In this question we have been asked to calculate the maximum possible velocity of the emitted electron. We have been given the cut-off frequency as well as radiation frequency. Therefore, we shall be using Einstein’s equation for photoelectric effect. This equation relates the kinetic energy of the photoelectron with total energy and the work function. Einstein stated that emission of photoelectrons is the result of interaction between single photons of incident radiation and electrons of metal.
Formula used:
\[K=E-\phi \]
Where,
K is the kinetic energy
E is the total energy
\[\phi \] is the energy required to remove electrons or work function.
Complete step-by-step answer:
According to Einstein the energy of the photon will be the sum of energy needed to remove electron and kinetic energy. This relation when stated for kinetic energy can be written as,
\[K=E-\phi \] ………….. (A)
Now, we know that radiation of frequency is given as 2v
Therefore, we can say that,
\[E=2hv\] …………………. (1)
Also, kinetic energy is given by,
\[K=\dfrac{1}{2}m{{V}^{2}}\] ……………….. (2)
It is given that the cut-off frequency is v. Therefore, energy required to remove electrons will be v.
Therefore, we can say that
\[\phi =hv\] ………………. (3)
Now, substituting the values from (1), (2) and (3) in equation (A)
We get,
\[\dfrac{1}{2}m{{V}^{2}}=2hv-hv\]
On solving,
We get,
\[{{V}^{2}}=\dfrac{2hv}{m}\]
Therefore,
\[V=\sqrt{\dfrac{2hv}{m}}\]
So, the correct answer is “Option C”.
Note: When light falls on the metal surface, the free electrons in the metal absorb this light and are liberated from the surface in the form of photons. This phenomenon is known as photoelectric emission. The law of photoelectric emission suggests that there is a definite cut-off frequency after which electrons can not be ejected and the number of emitted electrons is proportional to the intensity of incident light.
Formula used:
\[K=E-\phi \]
Where,
K is the kinetic energy
E is the total energy
\[\phi \] is the energy required to remove electrons or work function.
Complete step-by-step answer:
According to Einstein the energy of the photon will be the sum of energy needed to remove electron and kinetic energy. This relation when stated for kinetic energy can be written as,
\[K=E-\phi \] ………….. (A)
Now, we know that radiation of frequency is given as 2v
Therefore, we can say that,
\[E=2hv\] …………………. (1)
Also, kinetic energy is given by,
\[K=\dfrac{1}{2}m{{V}^{2}}\] ……………….. (2)
It is given that the cut-off frequency is v. Therefore, energy required to remove electrons will be v.
Therefore, we can say that
\[\phi =hv\] ………………. (3)
Now, substituting the values from (1), (2) and (3) in equation (A)
We get,
\[\dfrac{1}{2}m{{V}^{2}}=2hv-hv\]
On solving,
We get,
\[{{V}^{2}}=\dfrac{2hv}{m}\]
Therefore,
\[V=\sqrt{\dfrac{2hv}{m}}\]
So, the correct answer is “Option C”.
Note: When light falls on the metal surface, the free electrons in the metal absorb this light and are liberated from the surface in the form of photons. This phenomenon is known as photoelectric emission. The law of photoelectric emission suggests that there is a definite cut-off frequency after which electrons can not be ejected and the number of emitted electrons is proportional to the intensity of incident light.
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