The threshold wavelength for photoelectric emission from a metallic surface is 500\text{ n… — JAMB UTME

The threshold wavelength for photoelectric emission from a metallic surface is $500\text{ nm}$ . What is the work function of the metal in electron-volts ( $\text{eV}$ )? [Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^{8}\text{ m s}^{-1}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ ]

Answer: 2.48 eV

Answer

The work function of the metal is

2.48\text{ eV}

(or

2.475\text{ eV}

The work function

W_0

is calculated using

W_0 = \frac{hc}{\lambda_0}

. Substituting

h = 6.6 \times 10^{-34}\text{ J s}

c = 3.0 \times 10^8\text{ m s}^{-1}

, and

\lambda_0 = 500 \times 10^{-9}\text{ m}

gives

W_0 = 3.96 \times 10^{-19}\text{ J}

. Converting to electron-volts yields

\frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.475\text{ eV}

, which rounds to

2.48\text{ eV}

Step-by-Step Solution

State the relationship between work function and threshold wavelength

W_0 = \frac{hc}{\lambda_0}

The work function is the minimum energy required to liberate an electron, which corresponds to the maximum wavelength (threshold wavelength

\lambda_0

) that can cause emission.

Substitute the physical constants and threshold wavelength to calculate

W_0

in joules

W_0 = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{500 \times 10^{-9}} = 3.96 \times 10^{-19}\text{ J}

Evaluating

hc / \lambda_0

yields the work function energy in standard SI units (Joules).

Convert the calculated work function into electron-volts

W_0 = \frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.475\text{ eV}

Dividing the energy in Joules by

1.6 \times 10^{-19}\text{ J/eV}

converts the value to electron-volts.

Key Concept

Work Function and Threshold Wavelength Relationship

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