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Question: The work function of a metallic surface is \(5.01eV\). The photoelectrons are emitted when light of ...

The work function of a metallic surface is 5.01eV5.01eV. The photoelectrons are emitted when light of wavelength 2000A2000A falls on it. The potential difference applied to stop the fastest electrons is: [h=4.14×10−15eV sh=4.14\times {{10}^{-15}}eV\,s].
(A). 1.2V1.2V
(B). 2.4V2.4V
(C). 3.6V3.6V
(D). 4.8V4.8V

Explanation

Solution

When light falls on a metal surface, it ejects electrons from it due to the energy possessed by the photons. The minimum energy required to eject electrons is its work function. Using Einstein’s equation for photoelectric effect and substituting the corresponding values in it, we calculate the potential difference required to stop the electron with maximum kinetic energy or the fastest electron.

Formula used:
ν=cλ\nu =\dfrac{c}{\lambda }
eV0=hcλ−We{{V}_{0}}=\dfrac{hc}{\lambda }-W

Complete answer:
When light falls on a metal, the photons of light possess energy so when they collide with electrons they transfer their energy to them. If the energy transferred is equal to or more than the work function of the metal, then the electrons eject out.
The lowest energy required to eject an electron from a metal surface is known as the work function of the metal.
The equation for photoelectric effect given by Einstein is-
Kmax⁡=hν−W{{K}_{\max }}=h\nu -W
Here, Kmax⁡{{K}_{\max }} is the highest kinetic energy possessed by an electron of a metal in eV
hh is Planck’s constant
ν\nu is the frequency of light
WW is the value work function of the metal calculated in eV
We know that,
ν=cλ\nu =\dfrac{c}{\lambda }
Here, cc is the speed of light
λ\lambda is the wavelength
Substituting in the above equation, we get,
eV0=hcλ−We{{V}_{0}}=\dfrac{hc}{\lambda }-W - (1)
Here, V0{{V}_{0}} is the maximum potential difference applied at the ends of metal to accelerate to accelerate the electrons.
Given, e=1.6×10−19e=1.6\times {{10}^{-19}}, λ=2000×10−10m\lambda =2000\times {{10}^{-10}}m, W=5.01eV=5.01×1.6×10−19JW=5.01eV=5.01\times 1.6\times {{10}^{-19}}J, h=4.14×10−15eV s=4.14×10−15×1.6×10−19J sh=4.14\times {{10}^{-15}}eV\,s=4.14\times {{10}^{-15}}\times 1.6\times {{10}^{-19}}J\,s
Substituting given values in eq (1), we get,
1.6×10−19V0=4.14×10−15×1.6×10−19×3×1082000×10−10−5.01×1.6×10−19 ⇒V0=4.14×10−15×3×1082000×10−10−5.01 ⇒V0=6.21−5.01 ∴V0=1.2V \begin{aligned} & 1.6\times {{10}^{-19}}{{V}_{0}}=\dfrac{4.14\times {{10}^{-15}}\times 1.6\times {{10}^{-19}}\times 3\times {{10}^{8}}}{2000\times {{10}^{-10}}}-5.01\times 1.6\times {{10}^{-19}} \\\ & \Rightarrow {{V}_{0}}=\dfrac{4.14\times {{10}^{-15}}\times 3\times {{10}^{8}}}{2000\times {{10}^{-10}}}-5.01 \\\ & \Rightarrow {{V}_{0}}=6.21-5.01 \\\ & \therefore {{V}_{0}}=1.2V \\\ \end{aligned}
Therefore, the potential difference applied to stop the fastest electrons is 1.2V1.2V.

Hence, the correct option is (A).

Note:
The electrons which receive energy more than the work function are accelerated by the potential difference. The energy of the photons does not depend on the intensity but on the wavelength of the light. If we increase the intensity, the number of electrons ejected will increase. Reverse potential is applied to stop the motion of electrons.