Question

When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is –V₀/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is –V₀. The threshold frequency for photoelectric emission is :
(A) 5v/3
(B) 4v/3
(C) 2 v (D) 3v/2

Answer 1

Answer:

The threshold frequency of the light is given as

$\nu_{th} = \frac{3\nu}{2}$

Explanation:

As we know that the stopping potential of the electron is given as

$E = E_{th} + KE$

now we have

since we know that stopping potential is -Vo/2 so we have kinetic energy given as

$KE = \frac{eV_o}{2}$

$h\nu = E_{th} + \frac{eV_o}{2}$

Now the frequency of photon is changed to half

$h\frac{\nu}{2} = E_{th} + eV_o$

now from above 2 equations we have

$2h\nu - \frac{h\nu}{2} = E_{th}$

so we have

$h\nu_{th} = \frac{3h\nu}{2}$

so we have

$\nu_{th} = \frac{3\nu}{2}$

#Learn

Topic : Photo electric effect

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