Graph betwen stopping potential for most energetic emitted photoelectrons (V_ s ) with frequency (v) of incident radiation on metal is given below. Value of AB / BC , in graph is [where h= plank's constant, e=

Correct answer: C. h / e By Einstein's photoelectric equation aligned & K E_ =e V_s=h v-h v_0 & V_s=((h)/(e))v-(h v_0)/(e) aligned Graph of V_s with (v) is straight line whose slope (=(h)/(e)) Slope of graph (=(A B)/(B C) (A B)/(B C)=(h)/(e))

BITSAT 2014

Exam: BITSAT
Year: 2014
Subject: Physics
Chapter: Dual Nature of Matter

BITSAT 2014 · Physics · Dual Nature of Matter

Graph betwen stopping potential for most energetic emitted...

Graph betwen stopping potential for most energetic emitted photoelectrons

\left(\mathrm{V}_{\mathrm{s}}\right)

with frequency (v) of incident radiation on metal is given below. Value of

\mathrm{AB} / \mathrm{BC}

, in graph is [where

\mathrm{h}=

plank's constant,

\mathrm{e}=

electronic charge] Graph betwen stopping potential for most energetic emitted photoelectrons (V_ s ) with frequency (v) of incident

Options

A $h$
Be
C $\mathrm{h} / \mathrm{e}$
D $\mathrm{e} / \mathrm{h}$

Correct answer

C. $\mathrm{h} / \mathrm{e}$

Step-by-step solution

By Einstein's photoelectric equation

\begin{aligned}<br />& K E_{\max }=e V_s=h v-h v_0 \\<br />& \Rightarrow \quad V_s=\left(\frac{h}{e}\right)v-\frac{h v_0}{e}<br />\end{aligned}

Graph

of

V_s

with (\mathrm{v}) is straight line whose slope (=\frac{h}{e})
Slope of graph (=\frac{A B}{B C} \Rightarrow \frac{A B}{B C}=\frac{h}{e})

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