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Question: A light of wavelength \(5000A^\circ \) falls on a sensitive plate with work function \(1.7eV\). Find...

A light of wavelength 5000A∘5000A^\circ falls on a sensitive plate with work function 1.7eV1.7eV. Find the
(a) Energy of photon
(b) Energy of photoelectrons
(c) Stopping potential

Explanation

Solution

In this question we are provided with a light with some wavelength and we have to find the energy of the photon, energy of the photoelectron and the stopping potential. Then we are using the formula which is given below in the formula used and with substituting the given values we can find our solution.

Formula used:
Energy of a photon E=hfE = hf
where h is the planck's constant and f is the frequency of the photon
K.E=E−ϕK.E = E - \phi
where EE is the energy of photon and ϕ\phi is the work function
K.E=eV0K.E = e{V_0}
where V0{V_0} is the stopping potential

Complete step by step solution:
(a) ENERGY OF PHOTON (E)\left( E \right): E=hυE = h\upsilon
E=hcλE = h\dfrac{c}{\lambda } (∵υ=cλ)\left( {\because \upsilon = \dfrac{c}{\lambda }} \right) K.E=E−ϕK.E = E - \phi
We had wavelength (λ)\left( \lambda \right) be
\lambda = 5000A^\circ \\\ \Rightarrow \lambda = 5000 \times {10^{ - 10}}m \\\
planck's constant h=6.634×10−34Jsh = 6.634 \times {10^{ - 34}}Js and speed of light be c=3×108ms−2c = 3 \times {10^8}m{s^{ - 2}}
Putting the values
E=6.634×10−34×3×1085000×10−10JE = \dfrac{{6.634 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5000 \times {{10}^{ - 10}}}}J
Now converting the energy into eVeV by multiplying it with charge of the electron 1.6×10−191.6 \times {10^{ - 19}}
E=3.98×10−13J ⇒E=3.98×10−131.6×10−19eV ⇒E=2.48eV E = 3.98 \times {10^{ - 13}}J \\\ \Rightarrow E = \dfrac{{3.98 \times {{10}^{ - 13}}}}{{1.6 \times {{10}^{ - 19}}}}eV \\\ \Rightarrow E = 2.48eV

(b) ENERGY OF PHOTOELECTRON (K.E)\left( {K.E} \right)
K.E=E−ϕK.E = E - \phi
where ϕ\phi is the work function which is already given in the question to be 1.7eV1.7eV
K.E=(2.48−1.7)eV ⇒K.E=0.78eV K.E = \left( {2.48 - 1.7} \right)eV \\\ \Rightarrow K.E = 0.78eV

(c) STOPPING POTENTIAL (V0)\left( {{V_0}} \right)
Using the formula K.E=eV0K.E = e{V_0}
Now, substituting the above values
0.78eV=eV0 ⇒V0=0.78eVe ∴V0=0.78V 0.78eV = e{V_0} \\\ \Rightarrow {V_0} = \dfrac{{0.78eV}}{e} \\\ \therefore{V_0} = 0.78V

Note: When we are working with small systems then eVeV is so useful. Stopping potential is the potential need to stop the electron from removing when the incident photon has more energy than the work function of the metal. Work function is defined as the minimum work required to remove the electron from the solid.