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Question

A metallic surface is irradiated by a monochromatic light of frequency $$\upsilon_1$$ and stopping potential is found to be $$V_1$$. If the light of frequency $$\upsilon_2$$ irradiates the surface, the stopping potential will be


A
V1+he(υ1+υ2)
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B
V1+he(υ1υ2)
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C
V1+eh(υ2+υ1)
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D
V1he(υ1+υ2)
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Solution

The correct option is D $$V_1\, + \, \frac{h}{e}\left ( \upsilon _1\, - \, \upsilon _2 \right )$$
Maximum kinetic energy 
$$K_(max) = \dfrac{1}{2}mv^2 = ev_0$$
where $$ev_0$$ is the stopping potential. 
According to Einstein's photoelectric equation 
     $$h\upsilon_1 = \phi_0 + eV_1$$
     $$h\upsilon_2 = \phi_0 + eV_2$$
     $$\therefore h (\upsilon_1 - \upsilon_2) = e (V_1 - V_2)$$
    $$\dfrac{h}{e}(\upsilon_1 - \upsilon_2) = V_1 - V_2 $$
$$or V_2 = V_1 + \dfrac{h}{e}(\upsilon_2 - \upsilon_1)$$

Physics

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