MHT CET · Physics · Dual Nature of Matter
In a photoelectric experiment, a graph of maximum kinetic energy \(\left(\mathrm{KE}_{\max }\right)\) against the frequency of incident radiation \((v)\) is plotted. If \(\mathrm{A}\) and \(\mathrm{B}\) are the intercepts on the \(\mathrm{X}\) and \(\mathrm{Y}\) axis respectively the Planck's constant is given by
- A A+B
- B \(\frac{\mathrm{B}}{\mathrm{A}}\)
- C \(A \times B\)
- D \(\frac{\mathrm{A}}{\mathrm{B}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{B}}{\mathrm{A}}\)
Step-by-step Solution
Detailed explanation
Kinetic energy is given by
\(\text { (K.E.) })_{\max }=h v-h v_0\)
\(\text { Comparing }\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
We get \(\mathrm{x}\)-intercept when
\(\text { (K.E. })_{\max }=0 \text {, i.e., } \mathrm{h} v-\mathrm{h} v_0=0\)
Or \(v=v_0=\mathrm{A}\)
We get y-intercept when \(v=0\)
\(\therefore(\text { K.E. })_{\max }=-\mathrm{h} v_0=\mathrm{B}\)
\(\therefore\left|\frac{\mathrm{B}}{\mathrm{A}}\right|=\frac{\mathrm{h} \mathrm{v}_0}{\mathrm{v}_0}=\mathrm{h}\)
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