Example from Physics, 21.2 Einstein and the Photoelectric Effect

Photon Energy and the Photoelectric Effect: A Violet Light

(a) What is the energy in joules and electron volts of a photon of 420-nm violet light? (b) What is the maximum kinetic energy of electrons ejected from calcium by 420 nm violet light, given that the binding energy of electrons for calcium metal is 2.71 eV?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
To solve part (a), note that the energy of a photon is given by $E = h f$ . For part (b), once the energy of the photon is calculated, it is a straightforward application of $K E_{e} = h f - B E$ to find the ejected electron’s maximum kinetic energy, since BE is given.
Solution
Solution for (a)

Photon energy is given by

$E = h f .$

Since we are given the wavelength rather than the frequency, we solve the familiar relationship $c = f λ$ for the frequency, yielding

$f = \frac{c}{λ}$
21.7

Combining these two equations gives the useful relationship

$E = \frac{h c}{λ} .$
21.8

Now substituting known values yields

$E = \frac{(6.63 \times 10^{- 34} J \cdot s) (3.00 \times 10^{8} m / s)}{4.20 \times 10^{- 7} m} = 4.74 \times 10^{−19} J .$
21.9

Converting to eV, the energy of the photon is

$E = (4.74 \times 10^{- 19} J \cdot s) \frac{1 e V}{1.60 \times 10^{- 19} J} = 2.96 e V .$
21.10
Solution
Solution for (b)

Finding the kinetic energy of the ejected electron is now a simple application of the equation $K E_{e} = h f - B E$ . Substituting the photon energy and binding energy yields

$K E_{e} = h f - B E = 2.96 e V - 2.71 e V = 0.25 e V .$
21.11

0 of 3 steps open

How did it go?

Strategy

Solution

Solution

Strategy

Solution

Solution