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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Photoelectric Effect and Wave–Particle Duality

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Photoelectric Effect and Wave–Particle Duality

The figure displays the plot of the stopping potential versus the...

The figure displays the plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect. Determine the ratio h/eh/e.
The figure displays the plot of the stopping potential versus the frequency of the light used in an experiment on the

Options

  1. A8.28×10−15 V s8.28 \times 10^{-15} \text{ V s}
  2. B1.656×10−15 V s1.656 \times 10^{-15} \text{ V s}
  3. C4.14×10−15 V s4.14 \times 10^{-15} \text{ V s}
  4. D4.14×10−14 V s4.14 \times 10^{-14} \text{ V s}

Correct answer

C. 4.14×10−15 V s4.14 \times 10^{-15} \text{ V s}

Step-by-step solution

According to Einstein's photoelectric equation,

eV=hν−ΦeV = h\nu - \Phi

V=(he)ν−ΦeV = \left(\dfrac{h}{e}\right)\nu - \dfrac{\Phi}{e}

The graph of stopping potential VV versus frequency ν\nu is a straight line with slope equal to he\dfrac{h}{e}.

From the given graph, we can take two points: (ν1,V1)=(1×1014 Hz,0 V)(\nu_1, V_1) = (1 \times 10^{14} \text{ Hz}, 0 \text{ V}) and (ν2,V2)=(5×1014 Hz,1.656 V)(\nu_2, V_2) = (5 \times 10^{14} \text{ Hz}, 1.656 \text{ V}).

Slope =V2−V1ν2−ν1= \dfrac{V_2 - V_1}{\nu_2 - \nu_1}

he=1.656−0(5−1)×1014\dfrac{h}{e} = \dfrac{1.656 - 0}{(5 - 1) \times 10^{14}}

he=1.6564×1014\dfrac{h}{e} = \dfrac{1.656}{4 \times 10^{14}}

he=0.414×10−14 V s\dfrac{h}{e} = 0.414 \times 10^{-14} \text{ V s}

he=4.14×10−15 V s\dfrac{h}{e} = 4.14 \times 10^{-15} \text{ V s}
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