Question

The work function of caesium is 2.14 eV, find the wavelength of the incident light if photocurrent is brought to zero by stopping potential of 0.60 V.
(Ans : 4537 $\overset{\circ}{A}$ )

Answer 1

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I know the formula ; hf=work function+Ve where f=C/wavelength but it is not giving the desired answer maybe there's a mistake in my calculation. Try yourself

Answer 2

Answer:

4537 Angstrom

Explanation:

Given The work function of caesium is 2.14 eV, find the wavelength of the incident light if photo current is brought to zero by stopping potential of 0.60 V.

We know the relation hf = K max + C1 where photon energy is h f, k max is kinetic energy of the emitted electrons and work function is C1.  

So K max = e V

Now h f = e V + C1

So frequency of incident light is f = c / λ

  h c / λ = e V + C1

   h c / λ – C1 = e V

  – e V = C1 – h c / λ

   A λ = h c / e V + C1  

Substituting the values we get

λ = 6.62 x 10^-34 x 2.99 x 10^8 / 0.60 e V + 2.14 e V

  = 19.864748 x 10^-26 / 274 x 1.6 x 10^-19

 = 453.7 x 10 ^-9 n m

λ = 4537 Angstrom