---
title: "Photoelectric effect: work function of sodium"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:photoelectric-sodium-work-function"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/photoelectric-sodium-work-function
site: BCT Engineering
---

# Photoelectric effect: work function of sodium

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2082 Kartik (new course), 3 marks; 2074 Asoj, 4 marks (asked 2 times)
- Video: [Physical Sciences 2020:  Photoelectric Effect, Work Function, Kinetic Energy, Threshold Frequency](https://www.youtube.com/watch?v=-z-gvo30gP0) — Mindset
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/photoelectric-sodium-work-function
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

In the photoelectric experiment, green light, with a wavelength of 522 nm is the longest wavelength radiation that can cause photoemission of electron from a clean sodium surface. Calculate the work function of sodium. If ultraviolet radiation with a wavelength 250 nm is incident to the sodium surface, what will be the kinetic energy of the photo-emitted electrons?

## Short answer

Work function φ = hc/λ₀ = (6.626×10⁻³⁴×3×10⁸)/(522×10⁻⁹) = 3.81×10⁻¹⁹ J = 2.38 eV. Photon energy for 250 nm: E = 7.95×10⁻¹⁹ J = 4.97 eV. KE = E − φ = 4.14×10⁻¹⁹ J = 2.59 eV.

## Answer

In the photoelectric effect, a photon of energy $hf = hc/\lambda$ frees an electron only if $hf \ge \phi$ (work function). Einstein's equation is $KE_{max} = hf - \phi$. The **longest** wavelength that can cause emission is the threshold wavelength $\lambda_0$, for which $KE = 0$.

**Given:** $\lambda_0 = 522$ nm, $\lambda = 250$ nm, $h = 6.626\times10^{-34}$ J s, $c = 3\times10^{8}$ m/s, $1\text{ eV} = 1.6\times10^{-19}$ J.

#### Work function of sodium

$$
\begin{aligned}
\phi &= \frac{hc}{\lambda_0} = \frac{6.626\times10^{-34}\times3\times10^{8}}{522\times10^{-9}} \\
&= 3.808\times10^{-19}\ \text{J} = \frac{3.808\times10^{-19}}{1.6\times10^{-19}} = 2.38\ \text{eV}
\end{aligned}
$$

#### Energy of the UV photon

$$
\begin{aligned}
E &= \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\times3\times10^{8}}{250\times10^{-9}} \\
&= 7.951\times10^{-19}\ \text{J} = 4.97\ \text{eV}
\end{aligned}
$$

#### Kinetic energy of photoelectrons

$$
\begin{aligned}
KE_{max} &= E - \phi = 4.97 - 2.38 = 2.59\ \text{eV} \\
&= 4.14\times10^{-19}\ \text{J}
\end{aligned}
$$

**Answer:** work function of sodium $\phi = 3.81\times10^{-19}$ J $= 2.38$ eV; maximum kinetic energy of photoelectrons with 250 nm UV $= 4.14\times10^{-19}$ J $= 2.59$ eV.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is 522 nm used to find the work function?
   A. It is the shortest wavelength that causes emission
   B. It is the longest wavelength causing emission, where KE = 0 so phi = hc/lambda0
   C. It gives the maximum KE
   D. It is the wavelength of the UV light
   Answer: B. At the threshold wavelength the photon energy just equals the work function. (Key point: Threshold wavelength)

2. What is the work function of sodium?
   A. 2.38 eV
   B. 4.97 eV
   C. 1.24 eV
   D. 5.22 eV
   Answer: A. phi = hc/lambda0 = 3.81e-19 J = 2.38 eV. (Key point: Work function value)

3. What is the maximum KE of electrons for 250 nm light?
   A. 4.97 eV
   B. 7.35 eV
   C. 2.38 eV
   D. 2.59 eV
   Answer: D. KE = hf - phi = 4.97 - 2.38 = 2.59 eV. (Key point: Photoelectron KE)

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