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]

Question

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

Quantrex Academy · Accepted Answer

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))

Graph betwen stopping potential for most energetic emitted...

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