Question

What is the stopping potential when the metal with work function 0.6ev is illuminated with the light of 2ev ? (1) 2.6ev. (2) 3.6 ev. (3) 0.8 ev (4) 1.4ev

Answer 1

Answer on Question 75368, Physics, Quantum Mechanics

Question:

What is the stopping potential when the metal with work function $0.6\,eV$ is illuminated with the light of $2.0\,eV$?

a) $2.6\,V$

b) $3.6\,V$

c) $0.8\,V$

d) $1.4\,V$

Solution:

Using the mathematical description of the photoelectric effect, we can write the maximum kinetic energy $E_{Kmax}$ of an emitted electron as follows:

$$E_{Kmax} = hf - \varphi,$$

here, $hf = 2.0\,eV$ is the energy of the incident photon, $\varphi = 0.6\,eV$ is the work function for the metal.

From the other hand, the maximum kinetic energy of an emitted electron can be determined from the stopping potential:

$$E_{Kmax} = eV_s,$$

here, $e$ is the charge of electron, $V_s$ is the stopping potential.

Finally, equating these two equations we can find the stopping potential:

$$eV_s = hf - \varphi,$$ $$V_s = \frac{hf - \varphi}{e} = \frac{2.0\,eV \cdot \frac{1.6 \cdot 10^{-19}\,J}{1\,eV} - 0.6\,eV \cdot \frac{1.6 \cdot 10^{-19}\,J}{1\,eV}}{1.6 \cdot 10^{-19}\,C} = 2.0\,V - 0.6\,V = 1.4\,V.$$

Answer:

d) $V_s = 1.4\,V$

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