Question

Light of wavelength 199 nm shines on a metal surface. 4.08 eV is required to eject an electron. What is the kinetic energy of

(a) the fastest and
(b) the slowest ejected electrons?
(c) What is the stopping potential for this situation?
(d) What is the cutoff wavelength for this metal?

Answer 1

a) = 6.23eV

b) = 0eV

c) = 6.23V

d) = 304.12nm

Step-by-step explanation:

λ = 199nm

E = 4.08eV = Φ

hf = K(max) + Φ

Φ = work function of the target material

hf = photon energy

K(max) = k.e of the metal

K(max) = [(6.626*10⁻³⁴) * (3.0*10⁸)] / 199*10⁻⁹

K(max) = 9.99*10⁻¹⁹J

1.602*10⁻¹⁹J = 1eV

9.99*10⁻¹⁹ = 6.23eV

K(max) = 6.23eV

b). The slowest moving electron has a kinetic energy of zero

c). The stopping potential is the potential difference required to stop the fastest electron.

eV = K(max)

V = K(max) / e

V = 6.23eV / e

V = 6.23V

d)

The cut- off wavelength occurs when the maximum K.E is zero.

Φ = hc / λ

λ₀ = hc / Φ

λ₀ = (6.626*10⁻³⁴ * 3.0*10⁸) / (4.08 * 1.602*10⁻¹⁹)

λ₀ = 1.9878*10⁻²⁵ / 6.5361*10⁻¹⁹

λ₀ = 3.04*10⁻⁷m

λ₀ = 304.12nm