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Question: Up to what potential \[V\] can a zinc ball (work function \[3.74\,{\text{eV}}\] ) removed from other...

Up to what potential VV can a zinc ball (work function 3.74 eV3.74\,{\text{eV}} ) removed from other bodies be charged by irradiating it with light of λ=200 nm\lambda = 200\,{\text{nm}} ?
(A) 2.5 V2.5\,{\text{V}}
(B) 1.8 V1.8\,{\text{V}}
(C) 2.2 V2.2\,{\text{V}}
(D) 3 V3\,{\text{V}}

Explanation

Solution

First of all, we will use the equation given by Einstein, for the photoelectric effect. We will substitute the values in that equation and convert the quantities into similar units. We will manipulate the equation accordingly, to obtain the result.

Complete step by step answer:
Given,
Work function is 3.74 eV3.74\,{\text{eV}} .
Wavelength of the incident radiation is:
λ=200 nm\lambda = 200\,{\text{nm}}
We are now required to find the stopping potential.
Again, we know:
1 nm=1×10−9 m1\,{\text{nm}} = 1 \times {10^{ - 9}}\,{\text{m}}
So, the wavelength can be re-written as:
λ=200×10−9 m\lambda = 200 \times {10^{ - 9}}\,{\text{m}}
Electron volt in terms of joules as:
1 eV=1.6×10−19 J1\,{\text{eV}} = 1.6 \times {10^{ - 19}}\,{\text{J}}
To calculate the stopping potential, we use Einstein’s equation of photoelectric effect:
eV0=hν−ϕe{V_0} = h\nu - \phi …… (1)
Where,
ee indicates the charge of an electron.
V0{V_0} indicates the stopping potential.
hh indicates Planck’s constant.
ν\nu indicates frequency of the incident light.
ϕ\phi indicates work function.
Again, we know the formula of frequency of wave:
ν=cλ\nu = \dfrac{c}{\lambda } …… (2)
Where,
ν\nu indicates frequency of the incident light.
cc indicates velocity of light.
λ\lambda indicates wavelength of the incident light.
Now, we substitute the equation (2) in equation (1), for frequency and we get:
eV0=hν−ϕ ⇒eV0=h×cλ−ϕ  e{V_0} = h\nu - \phi \\\ \Rightarrow e{V_0} = h \times \dfrac{c}{\lambda } - \phi \\\
Now, we substitute the required values in the above equation and we get:
eV0=h×cλ−ϕ ⇒1.6×10−19 C×V0=6.626×10−34 Js×3×108 ms−1200×10−9 m−3.7 eV ⇒1.6×10−19 C×V0=6.626×10−34 Js×3×108 ms−1200×10−9 m−3.7×1.6×10−19 J ⇒1.6×10−19×V0=9.94×10−19−5.92×10−19  e{V_0} = h \times \dfrac{c}{\lambda } - \phi \\\ \Rightarrow 1.6 \times {10^{ - 19}}\,{\text{C}} \times {V_0} = 6.626 \times {10^{ - 34}}\,{\text{Js}} \times \dfrac{{3 \times {{10}^8}\,{\text{m}}{{\text{s}}^{ - 1}}}}{{200 \times {{10}^{ - 9}}\,{\text{m}}}} - 3.7\,{\text{eV}} \\\ \Rightarrow 1.6 \times {10^{ - 19}}\,{\text{C}} \times {V_0} = 6.626 \times {10^{ - 34}}\,{\text{Js}} \times \dfrac{{3 \times {{10}^8}\,{\text{m}}{{\text{s}}^{ - 1}}}}{{200 \times {{10}^{ - 9}}\,{\text{m}}}} - 3.7 \times 1.6 \times {10^{ - 19}}\,{\text{J}} \\\ \Rightarrow 1.6 \times {10^{ - 19}} \times {V_0} = 9.94 \times {10^{ - 19}} - 5.92 \times {10^{ - 19}} \\\
We carry out further simplifications to calculate the potential:
1.6×10−19×V0=4.02×10−19 ⇒V0=4.02×10−191.6×10−19 V ∴V0=2.5 V  1.6 \times {10^{ - 19}} \times {V_0} = 4.02 \times {10^{ - 19}} \\\ \Rightarrow{V_0} = \dfrac{{4.02 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\,{\text{V}} \\\ \therefore{V_0} = 2.5\,{\text{V}} \\\
Hence, the potential is found to be 2.5 V2.5\,{\text{V}} and the correct option is (A).

Note: To solve this problem, you must be well versed with Einstein's equation of photoelectric effect. It is important to subtract the work function from the energy of the incident radiation in order to find the energy associated with the outgoing electron. You can also go for an alternative solution by using the Planck’s constant in terms of another unit like electron volt, which is 4.136×10−15 eVs4.136 \times {10^{ - 15}}\,{\text{eVs}}. If you use this unit, then you need not convert eV{\text{eV}} to J{\text{J}}, for the work function.