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The graph of V_o (stopping potential) vs v (frequency) is

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a straight line with slope $(\frac{ϕ_{o}}{e})$

a straight line with slope $(- \frac{h}{ν_{o}})$

a straight line with slope $(- \frac{h}{e})$

a straight line with slope $(\frac{h}{e})$

answer is D.

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Detailed Solution

Since ${eV}_{o} = hν - ϕ_{o}$ where V_o = max. potential

or $V_{o} = (\frac{h}{e}) ν - \frac{ϕ}{e}$

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