Photoelectrons may be emitted from sodium (Φ = 2.36 eV) even for light intensities as low as 10^-8 W/m^2. Calculate classically how much time the light must shine to produce a photoelectron of kinetic energy 1.00 eV. Strategy We will assume that all of the light is absorbed in the first layer of

Question

Photoelectrons may be emitted from sodium [latex](\phi=2.36 eV )[/latex] even for light intensities as low as [latex]10^{-8} W / m ^{2}[/latex]. Calculate classically how much time the light must shine to produce a photoelectron of kinetic energy 1.00 eV.

Strategy We will assume that all of the light is absorbed in the first layer of atoms in the surface. Then we calculate the number of sodium atoms per unit area in a layer one atom thick. We assume that each atom in a single atomic layer absorbs equal energy, but a single electron in each of these atoms receives all the energy. We then calculate how long it takes these electrons to attain the energy (2.36 eV = 1.00 eV + 3.36 eV) needed for the electron to escape.

Accepted Answer

We first find the number of Na atoms/volume:

[latex]\begin{aligned}\frac{ ext { Avogadro's number }}{ ext { Na gram molecular weight }} imes ext { density } & \&=\frac{ ext { number of } Na ext { atoms }}{ ext { volume }}\end{aligned}[/latex]

[latex]\frac{6.02 imes 10^{23} ext { atoms } / mol }{23 g / mol } imes 0.97 \frac{ g }{ cm ^{3}}[/latex]

[latex]=2.54 imes 10^{22} \frac{ ext { atoms }}{ cm ^{3}}=2.54 imes 10^{28} \frac{ ext { atoms }}{ m ^{3}}[/latex] (3.26)

To estimate the thickness of one layer of atoms, we assume a cubic structure.

[latex]\frac{1 ext { atom }}{d^{3}}=2.54 imes 10^{28} \frac{ ext { atoms }}{ m ^{3}}[/latex]

[latex]d=3.40 imes 10^{-10} m[/latex]

= thickness of one layer of sodium atoms

If all the light is absorbed in the first layer of atoms, the number of exposed atoms per [latex]m ^{2}[/latex] is

[latex]2.54 imes 10^{28} \frac{ ext { atoms }}{ m ^{3}} imes 3.40 imes 10^{-10} m =8.64 imes 10^{18} \frac{ ext { atoms }}{ m ^{2}}[/latex]

With the intensity of [latex]10^{-8} W / m ^{2}[/latex], each atom will receive energy at the rate of

[latex]\begin{aligned}1.00 imes 10^{-8} \frac{ W }{ m ^{2}} imes & \frac{1}{8.64 imes 10^{18} ext { atoms } / m ^{2}} \&=1.16 imes 10^{-27} W \&=1.16 imes 10^{-27} \frac{ J }{ s } imes \frac{1}{1.6 imes 10^{-19} J / eV } \&=7.25 imes 10^{-9} eV / s\end{aligned}[/latex]

If energy is absorbed at the rate of [latex]7.25 imes 10^{-9}[/latex] eV/s for a single electron, we can calculate the time t needed to absorb 3.36 eV:

[latex]t=\frac{3.36 eV }{7.25 imes 10^{-9} eV / s }=4.63 imes 10^{8} s =14.7 ext { years }[/latex]

Based on classical calculations, the time required to eject a photoelectron should be 15 years!

Question 3.10: Photoelectrons may be emitted from sodium (Φ = 2.36 eV) even...

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