What is quantum yield in the experiment in percentage? | TestHub

Passage / Comprehension

Consider a classic experiment on the photoelectric effect in which a monochromatic plane wave of light falls on a metal plate of area $66.3 cm^{2}$ at the angle $3 0^{\circ}$ as shown in the figure. Electric field in the light wave varies according to equation $E = E_{0} cos (ω t + ϕ_{0})$ , where amplitude is $E_{0} = 15 V / m$ , angular frequency is $ω = 9.5 \times 1 0^{15} rad / s$ and $ϕ_{0}$ is some initial phase angle. For a plane wave, the density u of electromagnetic energy is given by the equation $u = \frac{1}{2} ε_{0} E_{0}^{2} . [h = 6.63 \times 1 0^{- 34} SI, ϵ_{0} = 8.85 \times 1 0^{- 12} SI]$ \\

\\ Dependence of photoelectric current I on voltage V applied between the electrodes is shown by a graph.\\

What is quantum yield in the experiment in percentage?

Answer:

3.80

Solution:

Angle of incidence = 60°.

Intensity $I = \frac{1}{2} ϵ_{0} E_{0}^{2} c$ ; power on plate $P = I A cos 6 0^{\circ} = \frac{1}{2} \times 8.85 \times 1 0^{- 12} \times (15)^{2} \times 3 \times 1 0^{8} \times 66.3 \times 1 0^{- 4} \times \frac{1}{2} = 9.9 \times 1 0^{- 4} W$

Photon flux: $n_{p} = \frac{P}{h ν} = \frac{P \times 2 π}{hω} = \frac{9.9 \times 1 0 ^{- 4} \times 2 π}{6.63 \times 1 0 ^{- 34} \times 9.5 \times 1 0 ^{15}} = 1 \times 1 0^{15} photons/sec$

From the graph, saturation current = 6 µA: $n_{e} = \frac{6 \times 1 0 ^{- 6}}{1.6 \times 1 0 ^{- 19}} = 3.75 \times 1 0^{13} electrons/sec$

Quantum yield $= \frac{n _{e}}{n _{p}} \times 100% = 3.75%$

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