Monochromatic light falls on a metal surface that has a work function of 6.7 x 10-19 J. Each photon has an energy of 8.0 x 10-19 J. (a) What is the maximum kinetic energy (in J) of the photoelectrons emitted by the surface? (b) What is the energy of each photon in electron volts eV? (c) What is the frequency of each photon (in Hz)?

Correct (a) Ekmax =  1.3 · 10⁻¹⁹ J, (b) ε = 5 eV, (c) f = 1.21 · 10¹⁵ Hz

Explanation:

Given:

photon energy  ε = 8 · 10⁻¹⁹ J

work function for metal  Wf = 6.7 · 10⁻¹⁹ J

(a) Ekmax = ?

The Einstein equation for photo effect is:

ε = Wf + Ekmax  ⇒ Ekmax = ε - Wf

Ekmax = 8 · 10⁻¹⁹ - 6.7 · 10⁻¹⁹ = 1.3 · 10⁻¹⁹ J

Ekmax =  1.3 · 10⁻¹⁹ J

(b)  ε = ? eV

the relationship between 1 eV and 1 J is:

1 eV = 1.6 · 10⁻¹⁹ J

so the energy of one photon in eV equals:

ε = 8 · 10⁻¹⁹ J = 8 · 10⁻¹⁹ / 1.6 · 10⁻¹⁹ eV = 5 eV

ε = 5 eV

(c)  f = ?

the energy of a single photon is calculated according to the formula:

ε = h · f  

where is:

h = 6.62 ·  10⁻³⁴ J s    plank constant

and f  frequency of each photon (electromagnetic radiation)

f = ε / h = 8 · 10⁻¹⁹ J / 6.62 ·  10⁻³⁴ J s = 1.21 · 10¹⁵ s⁻¹ =  1.21 · 10¹⁵ Hz

f = 1.21 · 10¹⁵ Hz

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answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.

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