The potential difference that must be applied to stop the fastest photoelectrons

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

Options

(a) 2.4 V
(b) – 1.2 V
(c) – 2.4 V
(d) 1.2 V

Correct Answer:

1.2 V

Explanation:

Kₘₐₓ = hc / – W = hc / λ – 5.01
= 12375 / λ(in Å) – 5.01
12375 / 2000 – 5.01 = 6.1875 – 5.01 = 1.17775
= 1.2 V

admin:

Related Questions

  1. A change of 0.04 V takes place between the base and the emitter when an input
  2. A radioactive substance emits n beta particles in the first 2 s and 0.5 n beta
  3. A solid sphere of mass M, radius R and having moment of inertia about an axis passing
  4. If by mistake voltmeter is connected in series and ammeter in parallel then
  5. Two balls are dropped from same height at 1 second interval of time.