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Question: Calculate the stopping potential \({V_{s2}}\) (A). 4.47 (B). 3.16 (C). 2.76 (D). 5.28...

Calculate the stopping potential Vs2{V_{s2}}
(A). 4.47
(B). 3.16
(C). 2.76
(D). 5.28

Explanation

Solution

- Hint: The photoelectric is the phenomenon in which electrons are emitted from the metal surface when the light of sufficient frequency is incident upon. Therefore, it implies that kinetic energy of electrons increases with light intensity.

Complete step-by-step solution -
Using the Einstein’s equation for photoelectric effect:
In first cases,
eVs2=hcλ1−W..............(1)e{V_{s2}} = \dfrac{{hc}}{{{\lambda _1}}} - W..............(1)
Where W= work function of metal surface.
And in second cases,
eVs2=hcλ2−W..............(2)e{V_{s2}} = \dfrac{{hc}}{{{\lambda _2}}} - W..............(2)
From equation (1) and (2), we get
eVs2=hcλ2−hcλ1+eVs1 ⇒(6.62×10−34)(3×108)190×10−9−(6.62×10−34)(3×108)550×10−9+(1.6×10−19)(0.19) ⇒7.152×10−19J or eVs2=7.152×10−191.6×10−19=4.47V  e{V_{s2}} = \dfrac{{hc}}{{{\lambda _2}}} - \dfrac{{hc}}{{{\lambda _1}}} + e{V_{s1}} \\\ \Rightarrow \dfrac{{(6.62 \times {{10}^{ - 34}})(3 \times {{10}^8})}}{{190 \times {{10}^{ - 9}}}} - \dfrac{{(6.62 \times {{10}^{ - 34}})(3 \times {{10}^8})}}{{550 \times {{10}^{ - 9}}}} + (1.6 \times {10^{ - 19}})(0.19) \\\ \Rightarrow 7.152 \times {10^{ - 19}}J \\\ or \\\ e{V_{s2}} = \dfrac{{7.152 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 4.47V \\\
Therefore the option (A) is the correct answer.
The stopping potential is defined as the voltage difference required to stop electrons from moving between plates and creating a current in the photoelectric experiment. The relationship between stopping potential and frequency is that the stopping potential is found to be changing linearly with frequency of incident of light being more negative for high frequency. Moreover, an increase in frequency of incident of light increases the kinetic energy of the emitted electrons which is so greater regarding potential that it is required to stop them completely.
The best example of photoelectric effect is when light shines on a cathode plate, the emitted electrons hit the anode from the plate and create a current. A solar panel is created from linking these together.

Note: Photoelectric effect is most commonly found in the solar panels. It is working on the basic principle of light striking the cathode which causes the emission of electrons, in turn which produces a current. Photoelectric effect is used in the form of photo-multiplier tubes and for charge coupled devices (CCDs).