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The maximum kinetic energy of the photoelectrons gets doubled when the wavelength

of light incident on the surface changes from lambda1 to lambda2. Derive the expressions for the

threshold wavelength lambda_{o} and work function for the metal surface.

we know that the maximum K.E of the electron ejected due to photoelectric effect is given by -

KEmax = hc/L - W,

where L is the wavelength of incident light and W is the work function. h is the planck constant and c is the speed of light in vacuum.

The work function is the property of the material from where the electron is ejected and represents the threshold energy that would be required to eject an electron so that it can escape the potential that binds all electrons.

So in the given problem we have,

k1 = hc/L1 - W

and

k2 = hc/L2 - W

, where L1 and L2 are incident photon wavelengths and k1 and k2 are maximum K.E. such that k2 = 2*k1.

Note since work function is the property of the material it will remain the same as the material on which the photon is incident is not changed.

Solving the above two equations we have -

k2 - k1=2*k1 - k1= k1= hc (1/L2 - 1/L1)

there fore we have expression for W as

W = hc/L1 - k1

on substituting the expression of k1

= hc (1/L1 - 1/L2 + 1/L1)

= hc (2/L1 - 1/L2)

Further to compute the threshold wavelength, L0 we have a condition

hc/L0 = W. Because at this wavelength the electron will just escape the binding potential but will have zero K.E.

Substituing the value of W we have

hc/L0 = hc (2/L1 - 1/L2)

therefore,

L0 = L1*L2/(2*L2 - L1)

Hope the concept clear now.

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