KCET · Physics · Dual Nature of Matter
In an experiment to study photoelectric effect the observed variation of stopping potential with frequency of incident radiation is as shown in the figure. The slope and \(y\)-intercept are respectively
- A \(\frac{h}{e},-\frac{h v_0}{e}\)
- B \(\frac{h v}{e}, v_0\)
- C \(\frac{h v}{e},-\frac{h}{e}\)
- D \(h v_1-h v_0\)
Answer & Solution
Correct Answer
(A) \(\frac{h}{e},-\frac{h v_0}{e}\)
Step-by-step Solution
Detailed explanation
The maximum kinetic energy of emitted
photoelectrons is given as
\(\begin{array}{ll}\Rightarrow & e V_0=h v-\phi_0 \\ \Rightarrow & V_0=\left(\frac{h}{e}\right) v-\frac{\phi_0}{e}\end{array}\)
Comparing with \(y=m x+c\), we get
\(\begin{aligned} \quad m & =\frac{h}{e} \text { and } c=-\frac{\phi_0}{e}=-\frac{h v_0}{e} \\ \therefore \text { Slope } & =m=\frac{h}{e}\end{aligned}\)
photoelectrons is given as
\(\begin{array}{ll}\Rightarrow & e V_0=h v-\phi_0 \\ \Rightarrow & V_0=\left(\frac{h}{e}\right) v-\frac{\phi_0}{e}\end{array}\)
Comparing with \(y=m x+c\), we get
\(\begin{aligned} \quad m & =\frac{h}{e} \text { and } c=-\frac{\phi_0}{e}=-\frac{h v_0}{e} \\ \therefore \text { Slope } & =m=\frac{h}{e}\end{aligned}\)
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