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Question: A metal surface of work function \(1.07\,eV\) is irradiated with light of wavelength \(332\,nm\). Th...

A metal surface of work function 1.07 eV1.07\,eV is irradiated with light of wavelength 332 nm332\,nm. The retarding potential required to stop the escape of photo electron is:
(A) 4.81 eV4.81\,eV
(B) 3.74 eV3.74\,eV
(C) 2.65 eV2.65\,eV
(D) 1.07 eV1.07\,eV

Explanation

Solution

The potential energy required to stop the escape of the photoelectron can be determined by using the energy of the incident light formula. Then the energy unit is converted into electron volt. By using the relation between the energy of the incident light and the work function, then the potential energy can be determined.

Formulae Used:
The energy of the incident light,
E=hcλE = \dfrac{{hc}}{\lambda }
Where, EE is the energy of the incident light, hh is the Planck’s constant and λ\lambda is the wavelength .

Complete step-by-step solution:
Given that,
The work function of the metal surface, W=1.07 eVW = 1.07\,eV
The wavelength of the light, λ=332 nm\lambda = 332\,nm
The energy of the incident light,
E=hcλ .................(1)E = \dfrac{{hc}}{\lambda }\,.................\left( 1 \right)
By substituting the Planck’s constant value, speed of light in free space vale and the wavelength of the light value in the above equation (1), then the above equation is written as,
E=6.626×10−34×3×108332×10−9E = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{332 \times {{10}^{ - 9}}}}
On multiplying the terms in the numerator, then the above equation is written as,
E=1.9878×10−25332×10−9E = \dfrac{{1.9878 \times {{10}^{ - 25}}}}{{332 \times {{10}^{ - 9}}}}
On dividing the above equation, then the above equation is written as,
E=5.98×10−19 JE = 5.98 \times {10^{ - 19}}\,J
By converting the unit of Energy from joules to electron volt, then the above equation is written as, ( 1 eV1\,eV is equal to the 1.6×10−19 J1.6 \times {10^{ - 19}}\,J, by using this the unit is converted)
E=3.73 eVE = 3.73\,eV
Then,
By photoelectric effect,
E=W0+eV0 ................(2)E = {W_0} + e{V_0}\,................\left( 2 \right)
Where, W0{W_0} is the work function and V0{V_0} is the potential difference.
By rearranging the above equation, then
V0=(E−W0)e{V_0} = \dfrac{{\left( {E - {W_0}} \right)}}{e}
By substituting the energy of the incident light and the work function values in the above equation, then
V0=3.73 eV−1.07 eVe{V_0} = \dfrac{{3.73\,eV - 1.07\,eV}}{e}
Taking the term ee as common in numerator, then
V0=(3.73−1.07) eVe{V_0} = \dfrac{{\left( {3.73 - 1.07} \right)\,eV}}{e}
By cancelling the same terms in numerator and denominator, then
V0=3.73−1.07 V{V_0} = 3.73 - 1.07\,V
By subtracting the terms in above equation, then
V0=2.66 V{V_0} = 2.66\,V
Thus, the above equation shows the potential required to stop the escape of the photoelectron.
Hence, the option (C) is the correct answer.

Note:- The work function is given in the unit of electron volt, so the energy of the incident light unit is converted from the joules to the electron volt. And then in equation (2), by keeping the potential in one side, then the potential in terms of volt is determined.