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Question 26

In a photoelectric effect experiment, the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be
Given E (in eV) = $$\frac{1237}{\lambda(\text{in nm})}$$

Solution

Step 1: Energy formula

$$E=\ \frac{\ 1237}{λ}$$

Step 2: Work function (threshold energy)

$$ϕ=\ \frac{\ 1237}{380}\approx3.26eV$$

Step 3: Incident photon energy

$$E=\ \frac{\ 1237}{260}\approx4.76eV$$

Step 4: Maximum kinetic energy

$$K_{\max}=E−ϕ=4.76−3.26=1.50eV$$

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