Question

What does it mean when we say that the $\beta $ -ray spectrum is a continuous energy spectrum?

Answer 1

Hint:The electromagnetic spectrum refers to the range of frequencies (the spectrum) of electromagnetic radiation, as well as the wavelengths and photon energies associated with each frequency. The electromagnetic waves within each frequency band are given different names, beginning at the low frequency (long wavelength) end of the spectrum with radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays, and ending with radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays at the high frequency. $\beta $ ray spectrum: The beta spectrum is the distribution of beta electron energy. It has the property that the emitting nucleus' kinetic energy is negligible, and the electron and antineutrino share the decay energy in varying proportions.

Complete answer:
A process in which the nucleus spontaneously releases an electron or a positron which is known as $\beta $− decay. The atomic number (Z) of the daughter nucleus is increased by unity and the mass (A) remains unchanged.
Symbolically $\beta $ − decay is represented as:
$_Z^AX \to _{Z + 1}^AY + _{ - 1}^0e + {v^ - }$
An example of $\beta $− decay is
 $_{90}^{234}Th \to _{91}^{234}Pa + _{ - 1}^0e + {v^ - }$
Here, we can see that charge and the number of nuclei are conserved in this type of decay.Antineutrino (${v^ - }$) is a neutral particle whose rest mass is very small. Total charge before reaction is +90e charge after reaction \[ - 91{\text{ }}e + \left( { - e} \right) + 090e\] .

Electrons and anti-neutrinos are neither nucleons nor anti-neutrinos. As a result, the number of nucleons remains constant at 234. Because it only includes neutrons and protons, the nucleus emits electrons (positrons) and neutrinos. We saw that an atom emits photons in the previous chapter, but we never say that photons are present in an atom. Photons are created during the emission process. Similarly, in the $\beta $ − decay, electrons (positrons) and antineutrino (neutrino) are formed. They are formed during the emission process.
We can also write.
$n \to p + {e^ - } + {v^ - }$ ……………. (1)
Here, we can see that nuclei do not vary in this process.

We can calculate disintegration energy in $\beta $ −decay. Let the nuclear masses of X and Y be ${m_x}\,and\,{m_y}$ and mass of electron is ${m_e}$ , then mass lost
$\Delta m = {m_x} - [{m_y} \times {m_e}]$ …(2)
Where antineutrino is considered to be massless. Adding and subtracting $Z{m_e}$ on the right side of equation (2), we get
$\Delta m = ({m_x} + Z{m_e}) - [{m_y} + (Z + 1){m_e}]$
$\Rightarrow \Delta m = {M_x} - {M_y}$ …(3)
Where ${M_x}\,and\,{M_y}$ the atomic masses of X and Y respectively, here,
Disintegration energy,
$Q = \Delta m{c^2}$
$\Rightarrow Q = ({M_x} - {M_y}){c^2}$ … (4)
The emitted $\beta $ − rays a range of energy varying from zero to a maximum value. The distribution of kinetic energies of $\beta $ −rays during $\beta $ −decay.

In positive $\beta $ (${\beta ^ + }$) decay, a positron and a neutrino are emitted from the nucleus. The symbolic representation of such a decay is
$_Z^AX \to _{Z - 1}^AY + _1^0{e^ + } + v$ ... (5)
An example of such a decay is:
$_{11}^{23}Na \to _{10}^{23}Ne + _1^0{e^ + } + v$
Here mass number A is unchanged whereas atomic reduces by unity. Like antineutrino, neutrino v is charged less and almost massless. Neutrino and positrons are also formed during this,
$n \to p + {e^ + } + v$ ... (6)
In ${\beta ^ + }$ decay, like ${\beta ^ - }$ decay, there is conservation of the number of nucleons and charge. The disintegration energy for ${\beta ^ + }$ decay,
$Q = ({M_x} - {M_y} - 2{m_e}){c^2}$
The parent nucleus accepts an electron, and the proton combines with it to form a neutron, which emits a neutrino in some P decays.

The following is a symbolic representation of such a decay:
$_Z^AX + _{ - 1}^0e \to _{Z - 1}^AY + v$
An example of such a process is:
$_{29}^{64}Cu + _{ - 1}^0e \to _{28}^{64}Ni + v$
and Q value for such a process,
$Q = ({M_x} - {M_y}){c^2}$
From $\beta $ -decay, we get to know that neutrons and protons are not fundamental particles. It is noteworthy that the process $n \to p + {e^ - } + v$ is possible both in the cases:inside and outside the nucleus, i.e. an isolated neutron can decay to form a proton.However the process $n \to p + {e^ + } + v$ is not possible outside the nucleus. As the mass of a neutron is more than that of protons. Thus, it is not possible to decay an isolated proton into a neutron.

Note:Electric and magnetic fields oscillate at right angles to each other to propagate electromagnetic waves. In vacuum, these waves travel at $3 \times {10^8}m{s^{ - 1}}$ the speed of light. Electric or magnetic fields have no effect on them. Interference or diffraction can be seen. They're known as transverse waves.
$Q = ({M_x} - {M_y}){c^2}$
From $\beta $-decay, we get to know that neutrons and protons are not fundamental particles. It is noteworthy that the process $n \to p + {e^ - } + v$ is possible both in the cases: inside and outside the nucleus, i.e. an isolated neutron can decay to form a proton. However the process $n \to p + {e^ + } + v$ is not possible outside the nucleus. As the mass of a neutron is more than that of protons. Thus, it is not possible to decay an isolated proton into a neutron.