The graph shows the variation of stopping potential V S with the frequency v of the incident light in a photoelectric experiment- From the graph- it can be chargeA- The work function of the metal is eb-B- The work function of the metal is ea-C- The value of Planck-s constant is equal ebeaD- The value of Planck-s constnat is equal to - b

The correct options are A The work function of the metal is eb. nbsp; C The value of Planck's constant is equal to ebaThe relation between the stopping potent ...

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Byju's Answer

Standard XII

Physics

Stopping Potential Revisited

The graph sho...

Question

The graph shows the variation of stopping potential $V_{S}$ with the frequency $v$ of the incident light in a photoelectric experiment. From the graph, it can be concluded that (e is electronic charge)

The work function of the metal is eb.

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The work function of the metal is ea.

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The value of Planck's constant is equal to

\frac{e b}{a}

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The value of Planck's constnat is equal to

\frac{e a}{b}

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Solution

The correct options are
A The work function of the metal is eb.
C The value of Planck's constant is equal to $\frac{e b}{a}$
The relation between the stopping potential and the frequency is
$h v = e V_{s} + ϕ$
$V_{s} = \frac{h v}{e} - \frac{ϕ}{e}$
From the graph, slope $= \frac{b}{a}$ .
$∴ \frac{b}{a} = \frac{h}{e} \Rightarrow h = \frac{e b}{a}$ .
From the graph, y intercept = b.
$∴ b = \frac{ϕ}{e} \Rightarrow ϕ = b e$ .

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Similar questions

Q. In a photocell circuit the stopping potential,

v_{0}

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e = 1.6 \times 10^{- 19} C

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Q. Plot a labelled graph of

| V_{s} |

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Q. The graph

\frac{1}{λ}

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(V)

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ϕ_{2}

and

ϕ_{3}

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(ii) Ratio of work functions

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The graph shows the variation of stopping potential VS with the frequency v of the incident light in a photoelectric experiment. From the graph, it can be concluded that (e is electronic charge)

The correct options are A The work function of the metal is eb. C The value of Planck's constant is equal to ebaThe relation between the stopping potential and the frequency is hv=eVs+ϕ Vs=hve−ϕe From the graph, slope =ba. ∴ba=he⇒h=eba. From the graph, y intercept = b. ∴b=ϕe⇒ϕ=be.

The graph shows the variation of stopping potential $V_{S}$ with the frequency $v$ of the incident light in a photoelectric experiment. From the graph, it can be concluded that (e is electronic charge)