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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
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The correct answer is:
$\frac{\mathrm{B}}{\mathrm{A}}$
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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