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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Atomic Physics

BITSAT 2022 · Physics · Atomic Physics

The stopping potential (V_0) versus frequency (v) of a graph for...

The stopping potential (V0)\left(V_0\right) versus frequency (v)(v) of a graph for photoelectric effect in a metal. From the graph, the planck's constant (h)(h) is. The stopping potential (V_0) versus frequency (v) of a graph for photoelectric effect in a metal. From the graph, the

Options

  1. A6.60×10−34J−s6.60 \times 10^{-34} \mathrm{ J}-\mathrm{s}
  2. B6.69×10−34J−s6.69 \times 10^{-34} \mathrm{ J}-\mathrm{s}
  3. C6.62×10−34J−s6.62 \times 10^{-34} \mathrm{ J}-\mathrm{s}
  4. D6.63×10−34J−s6.63 \times 10^{-34} \mathrm{ J}-\mathrm{s}

Correct answer

A. 6.60×10−34J−s6.60 \times 10^{-34} \mathrm{ J}-\mathrm{s}

Step-by-step solution

eV0=h(v−v0)=hv−hv0V0=(he)v−(he)v0…(i)\begin{aligned} e V_0=h\left( v-v_0\right) & =h v-h v_0 \\ V_0 & =\left(\frac{h}{e}\right) v-\left(\frac{h}{e}\right) v_0 \quad \quad \ldots (i) \end{aligned} From Eq.(i),itfollowsthattheslopeofthegraphisheEq. (i), it follows that the slope of the graph is \frac{h}{e}. Therefore, h=e× slope =1.6×10−19×165−0(8−4)×1015=1.6×10−19×16.54×1015=6.6×10−34J−s.\begin{aligned} h=e \times \text{ slope }& =1.6 \times10^{-19} \times \frac{165-0}{(8-4) \times 10^{15}}\\ & =\frac{1.6 \times 10^{-19} \times 16.5}{4 \times 10^{15}}=6.6 \times10^{-34} \mathrm{ J}-\mathrm{s} . \end{aligned}
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