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Graph of stopping potential for most ene...

Graph of stopping potential for most energetic emitted photoelectron `(V_(S))` with frequency of incident radiation on metal is given below.

The value of `(AB)/(BC)`, in graph is
(h=Planck's constant, e = electronic charge)

A

h

B

e

C

`(h)/(e)`

D

`(e)/(h)`

Text Solution

Verified by Experts

The correct Answer is:
C
By Einstein's photoelectric equation
`KE_("max")=eV_(s)=hv-hv_(0)`
`implies" "V_(s)=((h)/(e))v-(hv_(0))/(e)`
Graph of `V_(s)` with v is straight line whose slope = `(h)/(e)`
Slope of graph `=(AB)/(BC)implies (AB)/(BC)=(h)/(e)`
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Knowledge Check

  • 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 = electronic charge]

    A
    h
    B
    e
    C
    h/e
    D
    e/h

  • The kinetic energy (E_(k)) of a photoelectron varies with the frequency (v) of the incident radiation as which of the following graph ?

    A
    B
    C
    D

  • Which of the following is the graph between the frequency (v) of the incident radiations and the stopping potential ?

    A
    B
    C
    D

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