Photoelectric Calculations

In Lesson 4 you met Einstein's photoelectric equation:

hf = φ + (1/2)mv²_max

This relates four physical quantities — photon frequency f, work function φ, electron mass m and maximum photoelectron speed v_max — together with Planck's constant h. In OCR exam questions you will be asked to find any one of these four quantities, given the others, and you will also be asked about the threshold frequency, the threshold wavelength and the stopping potential V_s.

This lesson is a tour of the main calculation types you will encounter. Each example is worked out carefully, step by step, with particular attention to the unit conversions that cause so many careless errors.

The methods here are drawn directly from the OCR A-Level Physics A specification (H556), Module 4.5.

---

The Toolkit

Let us begin by listing every relevant formula:

Quantity	Formula
Photon energy	E_photon = hf = hc/λ
Einstein equation	hf = φ + KE_max
Max kinetic energy	KE_max = hf - φ
Max speed	v_max = √(2 KE_max/m)
Stopping potential	eV_s = KE_max
Threshold frequency	f₀ = φ/h
Threshold wavelength	λ₀ = c/f₀ = hc/φ

Constants:

h = 6.63 × 10⁻³⁴ J s
c = 3.00 × 10⁸ m s⁻¹
e = 1.60 × 10⁻¹⁹ C
m_e = 9.11 × 10⁻³¹ kg
1 eV = 1.60 × 10⁻¹⁹ J

As in the previous lesson, hc ≈ 1.99 × 10⁻²⁵ J m, or equivalently ≈ 1240 eV nm. These shortcuts save time and reduce errors.

---

Calculation 1: Finding the Work Function

When light of wavelength 350 nm falls on a metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.70 × 10⁻¹⁹ J. Calculate the work function of the metal.

Solution. Compute the photon energy:

E_photon = hc/λ = (1.99 × 10⁻²⁵)/(350 × 10⁻⁹) = 5.69 × 10⁻¹⁹ J

Rearrange Einstein's equation:

φ = E_photon - KE_max
  = (5.69 × 10⁻¹⁹) - (1.70 × 10⁻¹⁹)
  = 3.99 × 10⁻¹⁹ J

Convert to eV:

φ = (3.99 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹) ≈ 2.49 eV

Answer: φ ≈ 4.0 × 10⁻¹⁹ J ≈ 2.5 eV. (A value characteristic of potassium or sodium.)

---

Calculation 2: Threshold Frequency and Wavelength

A metal has work function φ = 4.50 eV. Calculate (a) the threshold frequency and (b) the threshold wavelength for photoemission from this metal.

Solution. Convert the work function to joules:

φ = (4.50)(1.60 × 10⁻¹⁹) = 7.20 × 10⁻¹⁹ J

(a) Threshold frequency:

f₀ = φ/h = (7.20 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 1.086 × 10¹⁵ Hz

(b) Threshold wavelength:

λ₀ = c/f₀ = (3.00 × 10⁸)/(1.086 × 10¹⁵) = 2.76 × 10⁻⁷ m = 276 nm

Or equivalently, using the shortcut hc ≈ 1240 eV nm:

λ₀ = hc/φ = 1240/4.50 ≈ 276 nm

Answer: f₀ ≈ 1.09 × 10¹⁵ Hz; λ₀ ≈ 276 nm.

This is in the far ultraviolet, which tells us the metal will not emit photoelectrons under visible light — probably a metal like zinc or aluminium.

---

Calculation 3: Maximum Photoelectron Speed

UV light of wavelength 250 nm falls on a potassium surface (work function 2.30 eV). Calculate the maximum speed of the emitted photoelectrons.

Solution. Work in eV where possible, then convert.

Photon energy: E_photon = hc/λ = 1240/250 = 4.96 eV.

KE_max = 4.96 - 2.30 = 2.66 eV.

Convert to joules:

KE_max = (2.66)(1.60 × 10⁻¹⁹) = 4.26 × 10⁻¹⁹ J

Apply the kinetic-energy formula:

(1/2)m v²_max = 4.26 × 10⁻¹⁹
v²_max = (2)(4.26 × 10⁻¹⁹)/(9.11 × 10⁻³¹)
      = 9.35 × 10¹¹
v_max = √(9.35 × 10¹¹) ≈ 9.67 × 10⁵ m s⁻¹

Answer: v_max ≈ 9.7 × 10⁵ m s⁻¹, or about 0.3% of the speed of light.

---

Calculation 4: Stopping Potential

Monochromatic light of wavelength 310 nm illuminates a metal surface whose work function is 3.00 eV. Calculate the stopping potential of the emitted photoelectrons.

Solution. The key relation is

eV_s = KE_max

Photon energy: E_photon = 1240/310 ≈ 4.00 eV.

KE_max = 4.00 - 3.00 = 1.00 eV.

Since eV_s = KE_max and electronvolts are particularly convenient for this calculation,

V_s = 1.00 V

Answer: V_s = 1.00 V.

Exam Tip: When you see an exam question asking for the stopping potential, and both the work function and photon energy are given (or easy to compute) in eV, the answer is just the difference in volts. No conversions to joules, no factors of e, just V_s = (E_photon in eV) - (φ in eV).

---

Calculation 5: Finding Planck's Constant from a Graph

In a photoelectric experiment, the following data are collected:

f / 10¹⁴ Hz	V_s / V
5.5	0.16
6.5	0.58
7.5	0.99
8.5	1.40
9.5	1.82

Use these values to determine Planck's constant experimentally.

Solution. Since eV_s = hf - φ, plotting V_s against f gives a straight line with slope h/e and y-intercept -φ/e.

Take two well-separated points: (5.5 × 10¹⁴, 0.16) and (9.5 × 10¹⁴, 1.82).

Gradient:

m = (1.82 - 0.16)/((9.5 - 5.5) × 10¹⁴)
  = 1.66 / (4.0 × 10¹⁴)
  = 4.15 × 10⁻¹⁵ V s

Hence:

h = (gradient) × e
  = (4.15 × 10⁻¹⁵)(1.60 × 10⁻¹⁹)
  = 6.64 × 10⁻³⁴ J s

Answer: h ≈ 6.6 × 10⁻³⁴ J s, in excellent agreement with the accepted value.

This kind of graphical data-handling question is a classic OCR format — it tests the Einstein equation, the stopping-potential relation, gradient analysis, and unit reasoning, all in one problem.

---

Calculation 6: Photocurrent and Intensity