Question

If the work function for certain metal is 1.8 eV, what is the stopping potential for electrons ejected from metal when light of 4000 $\overset{\circ}{A}$ shines on the metal ?
(Ans: 1.31 V)

Answer 1

hf= work function+ K.E

Where K.E is the kinetic energy and equal to Ve where V is the stopping potential and e is the charge on electron (1.6 × 10^-19)

Answer 2

Answer:

1.3 V

Explanation:

Given If the work function for certain metal is 1.8 eV, what is the stopping potential for electrons ejected from metal when light of 4000 ,\overset{\circ}{A}, shines on the metal ?

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

4000 Angstrom = 400 nanometre

So K max = e V

Now h f = e V + Ø

So frequency of incident light is f = c / λ

  h c / λ = e V + Ø

To solve for stopping potential we get

   h c / λ – eV = Ø

  – e V = Ø – h c / λ

  V = Ø / e – h c / e λ

V = h c / e λ – Ø / e

Substituting the value we get

V = 6.62 x 10^-34 x 2.99 x 10^8 / 1.602 x 10^-10 x 400 x 10^-9  - 1.8 V x e / e

3.1 V – 1.8 V

V = 1.3 V