Electrons are emitted from metals during photoelectric emission, thermionic emission and radioactive decay - Leaving Cert Physics - Question 8 - 2019

The electrons emitted during radioactive decay are referred to as beta (eta) particles.

A line emission spectrum is a series of discrete lines of color (frequencies) emitted by a material. It occurs when electrons in an atom absorb energy and move to higher energy levels. When these electrons return to their original states, they emit energy in the form of light at specific frequencies, corresponding to the energy differences between the levels.

Einstein's photoelectric equation is given by:
[ E_k = hf - \phi ]
where $E_k$ is the kinetic energy of the emitted electrons, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the material.

Point A represents the threshold frequency, which is the minimum frequency of incident radiation required to emit electrons from the metal surface.

The slope of the graph represents Planck's constant, $h$. It indicates the relationship between the kinetic energy of emitted electrons and the frequency of the incident radiation.

First, we calculate the kinetic energy of the emitted electrons:
[ E_k = E - \phi = 4.15 \text{ eV} - 3.68 \text{ eV} = 0.47 \text{ eV} ]
Next, we convert the kinetic energy into joules:
[ E_k = 0.47 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 7.5 \times 10^{-20} \text{ J} ]
Using the kinetic energy formula:
[ E_k = \frac{1}{2} mv^2 ]
we can find the maximum velocity:
[ v = \sqrt{\frac{2E_k}{m}} ]
Substituting in the values:
[ m = 9.1 \times 10^{-31} \text{ kg} ]
[ v = \sqrt{\frac{2 \times 7.5 \times 10^{-20}}{9.1 \times 10^{-31}}} = 4.1 \times 10^6 \text{ m/s} ]

Electrons are produced in the cathode region of the X-ray tube, where a material is heated to facilitate thermionic emission.

The minimum wavelength can be found using the equation:
[ E = \frac{hc}{\lambda} ]
Where $E$ is the energy, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
Converting 50 kV to joules:
[ E = 50 \times 10^3 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 8 \times 10^{-15} \text{ J} ]
Thus, using this energy value, we can find the minimum wavelength:
[ \lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{8 \times 10^{-15}} = 2.5 \times 10^{-11} \text{ m} ]
This corresponds to a minimum wavelength in X-rays.

1. Cooling system: X-ray tubes are often equipped with cooling systems to dissipate heat generated by the electrons when they strike the target.
2. Tungsten target: The target material is usually tungsten, which has a high melting point, allowing it to withstand the high temperatures produced during X-ray generation.