
Light of wavelength ‘A’ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential will:
(A) increase
(B) decrease
(C) be zero
(D) become exactly half
Answer
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Hint: In the above said, procedure the wavelength of light is reduced. Therefore, we shall see the relation between the dependence of stopping potential on the wavelength of incident photons. This relation is given by Einstein’s equation of photoelectric effect. From observing that equation under different wavelengths, we will be able to answer if the stopping potential will increase, decrease, or have other changes.
Complete answer:
The energy of incident photons can be written as:
$\Rightarrow E=\dfrac{hc}{A}$
Where,
(h) is the Planck’s constant.
And, (c) is the speed of light.
Now, the Einstein’s equation for photoelectric effect is given as:
$\Rightarrow E=K.E{{.}_{\max }}+\phi $
Here,
$\phi $ is the work function of the metal on which light is incident. This is the minimum energy required by the metal to just start the photoelectric effect. The work function of a metal is a constant value.
We can simplify Einstein’s equation as:
$\Rightarrow \dfrac{hc}{A}=e{{V}_{0}}+\phi $
Where,
${{V}_{0}}$is the stopping potential .
Now, from the above equation we can say that, the wavelength of incident light is inversely proportional to the stopping potential. Mathematically, this could be written as:
$\Rightarrow {{V}_{0}}\propto \dfrac{1}{A}$
Therefore, on decreasing the wavelength of the incident light, we are increasing the stopping potential.
Hence, the stopping potential will increase.
Hence, option (A) is the correct option.
Note:
We did not take the effect of changing wavelength on work function on metal because it is a constant value. It can be changed only by changing the physical conditions in which the metal is kept, that are pressure, temperature, etc. Also on decreasing the wavelength of incident light, the incident energy on the metal is increased, so the maximum kinetic energy of photoelectrons is also increased.
Complete answer:
The energy of incident photons can be written as:
$\Rightarrow E=\dfrac{hc}{A}$
Where,
(h) is the Planck’s constant.
And, (c) is the speed of light.
Now, the Einstein’s equation for photoelectric effect is given as:
$\Rightarrow E=K.E{{.}_{\max }}+\phi $
Here,
$\phi $ is the work function of the metal on which light is incident. This is the minimum energy required by the metal to just start the photoelectric effect. The work function of a metal is a constant value.
We can simplify Einstein’s equation as:
$\Rightarrow \dfrac{hc}{A}=e{{V}_{0}}+\phi $
Where,
${{V}_{0}}$is the stopping potential .
Now, from the above equation we can say that, the wavelength of incident light is inversely proportional to the stopping potential. Mathematically, this could be written as:
$\Rightarrow {{V}_{0}}\propto \dfrac{1}{A}$
Therefore, on decreasing the wavelength of the incident light, we are increasing the stopping potential.
Hence, the stopping potential will increase.
Hence, option (A) is the correct option.
Note:
We did not take the effect of changing wavelength on work function on metal because it is a constant value. It can be changed only by changing the physical conditions in which the metal is kept, that are pressure, temperature, etc. Also on decreasing the wavelength of incident light, the incident energy on the metal is increased, so the maximum kinetic energy of photoelectrons is also increased.
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