Photoelectric effect and quanta of radiation

a. The detachment of an electron from a metal requires an amount of energy characteristic of the metal, called the work function. The energy of light is absorbed into the metal one quantum at a time. The energy of a quantum is directly proportional to the frequency of light: $$ E=hf$$. A single quantum of visible light does not have enough energy to detach an electron, but the higher-frequency UV light quantum does.

b. Increasing the intensity does not change the energy of an individual quantum, which depends only on the frequency of the light. The energy of a visible light quantum is not sufficient to discharge the electric charge, so increasing the intensity does not change the situation.

a. In the photoelectric effect, the conservation of energy implies that the energy of a photon absorbed by the metal is converted into the work function and the kinetic energy of the emitted electron. This is described by the following equation:

$$ hf=W_0+E_{\text{k}}$$

When this is combined with the expression for the energy of a radiation quantum $$ E_{\text{quantum}}=hf$$, the equation of the straight line becomes:

$$ E_{\text{k}}=hf-<!--\; -\; -->W_0$$

b. Planck’s constant is the slope of the line. Based on the obtained parameters

$$ h=4.3\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->15}\text{ eVs}$$

The point where the line intersects the vertical axis is the negative value of the work function. Based on the parameters of the model

$$ W_0=2.0\text{ eV}$$

The metal could be caesium, as the work function is close to its tabulated value.

a. The energy of a photon can be calculated using the equation $$ E=hf$$. By combining it with the basic equation of wave motion, the formula can be expressed as $$ E={\displaystyle \frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}}$$. 

$$ E={\displaystyle \frac{6.626\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->34}\text{ Js}\cdot <!--\; \cdot \; -->2.998\cdot <!--\; \cdot \; -->{10}^8\text{ m/s}}{514\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->9}\text{ m}}}=3.86468\dots <!--\; \dots \; -->\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->19}\text{ J}\approx <!--\; \approx \; -->3.86\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->19}\text{ J}$$

$$ E={\displaystyle \frac{4.136\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->15}\text{ eVs}\cdot <!--\; \cdot \; -->2.998\cdot <!--\; \cdot \; -->{10}^8\text{ m/s}}{514\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->9}\text{ m}}}=2.412\dots <!--\; \dots \; -->\text{eV}\approx <!--\; \approx \; -->2.41\text{ eV}$$

The energy is 2.41 eV in electronvolts and 3.86 ⋅ 10⁻¹⁹ J in joules.

b. To solve the velocity, the kinetic energy is set equal to the energy of the photon. The energy of the photon must be expressed in joules in the calculation, as it is the derived unit for kinetic energy based on the base units of velocity and mass.

$$ {\displaystyle \frac{1}{2}}m{v}^2=E{}_{\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}}$$

$$ v=\sqrt{{\displaystyle \frac{2E{}_{\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}}}{m}}}$$

$$ v=\sqrt{{\displaystyle \frac{2\cdot <!--\; \cdot \; -->3.864\mathrm{68...}\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->19}\text{ J}}{9.109\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->31}\mathrm{k}\mathrm{g}}}}\approx <!--\; \approx \; -->920{\displaystyle \frac{\mathrm{k}\mathrm{m}}{\mathrm{s}}}$$

The speed is 920 km/s.

In the photoelectric effect, the energy of the photon absorbed by the metal is converted into the work function and the kinetic energy of the emitted electron:

$$ hf=W_0+E_{\text{k}}$$

The energy of a photon can be expressed using the wavelength. The electric field does the work $$ W=QU=eU=0.29\text{ eV}$$ which is equal to the kinetic energy of the electrons, causing them to stop. The following equation is obtained:

$$ {\displaystyle \frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}}=W_0+eU$$

From this, the work function can be solved as:

$$ W_0={\displaystyle \frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}}-<!--\; -\; -->eU={\displaystyle \frac{4.136\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->15}\mathrm{e}\mathrm{V}\mathrm{s}\cdot <!--\; \cdot \; -->2.998\cdot <!--\; \cdot \; -->{10}^8\mathrm{m}/\mathrm{s}}{614\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->9}\mathrm{m}}}-<!--\; -\; -->0.29\mathrm{e}\mathrm{V}=1.7327\dots <!--\; \dots \; -->\text{ eV}\approx <!--\; \approx \; -->1.7\text{ eV}$$

The conservation of energy equation can now be applied again to the situation where the wavelength is 468 nm. Solve for the work done by the field, $eU$:

$$ eU={\displaystyle \frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}}-<!--\; -\; -->W_0={\displaystyle \frac{4.136\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->15}\mathrm{e}\mathrm{V}\mathrm{s}\cdot <!--\; \cdot \; -->2.998\cdot <!--\; \cdot \; -->{10}^8\mathrm{m}/\mathrm{s}}{468\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->9}\mathrm{m}}}-<!--\; -\; -->1.733\text{ eV}=0.916\dots <!--\; \dots \; -->\text{eV}\approx <!--\; \approx \; -->0.92\text{ eV}$$

The field must do 0.92 eV of work, so a voltage of 0.92 volts is required.