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Exercises · 2.9

Q.A photon of wavelength 4×10−7 m4 \times 10^{-7}\ m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate

(i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and
(iii) the velocity of the photoelectron (1 eV=1.6020×10−19 J1\ eV = 1.6020 \times 10^{-19}\ J).
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The problem uses Einstein’s photoelectric equation: photon energy equals work function plus kinetic energy of the emitted electron. First, convert the photon’s wavelength to energy in joules, then to eV. Subtract the work function to get kinetic energy, and finally use K=12mv2K = \frac{1}{2} m v^2 to find the electron’s velocity.

Why this approach works

The photoelectric effect is a clean demonstration of light behaving as particles (photons). Each photon carries a discrete packet of energy given by E=hν=hcλE = h \nu = \frac{h c}{\lambda}. When this photon hits a metal surface, part of its energy is used to free the electron from the metal (the work function ϕ\phi), and the remainder appears as the electron’s kinetic energy. This is Einstein’s photoelectric equation:

Ephoton=ϕ+KmaxE_{\text{photon}} = \phi + K_{\text{max}}

where KmaxK_{\text{max}} is the maximum kinetic energy of the emitted photoelectron. From KmaxK_{\text{max}}, we can find the electron’s speed using K=12mev2K = \frac{1}{2} m_e v^2, since the electron’s mass is known.

Step-by-step solution

1. Find the photon energy in joules

The energy of a single photon is:

E=hcλE = \frac{h c}{\lambda}

where

h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34}\ \text{J·s} (Planck’s constant)

c=3.0×108 m/sc = 3.0 \times 10^{8}\ \text{m/s} (speed of light)

λ=4×10−7 m\lambda = 4 \times 10^{-7}\ \text{m}

Substitute:

E=(6.626×10−34)(3.0×108)4×10−7E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{4 \times 10^{-7}}

First compute numerator: 6.626×3.0=19.8786.626 \times 3.0 = 19.878, and 10−34×108=10−2610^{-34} \times 10^{8} = 10^{-26}, so numerator = 19.878×10−2619.878 \times 10^{-26}.

Divide by 4×10−74 \times 10^{-7}:

E=19.878×10−264×10−7=4.9695×10−19 JE = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}} = 4.9695 \times 10^{-19}\ \text{J}

Tip

A quick check: visible light photons have energies around 10−1910^{-19} J, so this result is in the right ballpark.

2. Convert photon energy to eV

Given 1 eV=1.6020×10−19 J1\ \text{eV} = 1.6020 \times 10^{-19}\ \text{J}:

E (eV)=4.9695×10−191.6020×10−19=3.10 eVE\ (\text{eV}) = \frac{4.9695 \times 10^{-19}}{1.6020 \times 10^{-19}} = 3.10\ \text{eV}

So the photon energy is 3.10 eV.

3. Calculate the kinetic energy of the emitted electron

Work function ϕ=2.13 eV\phi = 2.13\ \text{eV}. Using Einstein’s equation:

Kmax=Ephoton−ϕ=3.10 eV−2.13 eV=0.97 eVK_{\text{max}} = E_{\text{photon}} - \phi = 3.10\ \text{eV} - 2.13\ \text{eV} = 0.97\ \text{eV} …

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