When a certain photosensitive surface is illuminated with monochromatic light of frequency ν, the stopping potential for the photo current is -V02, When the surface is illuminated by monochromatic light of frequency ν2, the stopping potential is -V0. The threshold frequency for photoelectric emission is

Using to Einstein's Photoelectric Equation, we havehν=ϕ0+eV02 ⇒2hν=2ϕ0+eV0 ⋯ i Also, hν2=ϕ0+eV0 ⋯ ii Subtracting (ii) from (i), ⇒3hν2=ϕ0=hν0 ⇒ν0=3ν2

When a certain photosensitive surface is illuminated with monochromatic light of frequency ν, the stopping potential for the photo current is -V02, When the surface is illuminated by monochromatic light of frequency ν2, the stopping potential is -V0. The threshold frequency for photoelectric emission is

Using to Einstein's Photoelectric Equation, we havehν=ϕ0+eV02 ⇒2hν=2ϕ0+eV0 ⋯ i Also, hν2=ϕ0+eV0 ⋯ ii Subtracting (ii) from (i), ⇒3hν2=ϕ0=hν0 ⇒ν0=3ν2

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When a certain photosensitive surface is illuminated with monochromatic light of frequency $ν$ , the stopping potential for the photo current is $- \frac{V_{0}}{2}$ , When the surface is illuminated by monochromatic light of frequency $\frac{ν}{2}$ , the stopping potential is $- V_{0}$ . The threshold frequency for photoelectric emission is

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$2 ν$

$\frac{3 ν}{2}$

$\frac{4}{3} ν$

$\frac{5 ν}{3}$

answer is A.

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Detailed Solution

Using to Einstein's Photoelectric Equation, we have

$h ν = ϕ_{0} + e (\frac{V_{0}}{2}) \Rightarrow 2 h ν = 2 ϕ_{0} + e V_{0} \dots (i)$

$Also, \frac{h ν}{2} = ϕ_{0} + e V_{0} \dots (i i) Subtracting (ii) from (i), \Rightarrow \frac{3 h ν}{2} = ϕ_{0} = h ν_{0} \Rightarrow ν_{0} = \frac{3 ν}{2}$

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