Question

An alpha particle is accelerated through a potential difference of ${10^6}$ volt. Its kinetic energy will be:
A. $1\,MeV$
B. $2\,MeV$
C. $4\,MeV$
D. $8\,MeV$

Answer 1

Hint: An alpha particle is considered to be almost equivalent to the nucleus of a Helium particle. The formula for the kinetic energy in terms of the charge of a particle and the potential difference of a particle is applied. The charge of the alpha particle is determined and applied to the formula to find out the kinetic energy of the alpha particle.

Formula used:
The work done is given by:
$W = qV$
Where, $q$ is the charge and $V$ is the potential difference or the voltage.

Complete step by step answer:
The above problem revolves around the concept of the kinetic energy of the alpha particles present in the radiation, that is, with what energy the alpha particles are travelling with. In order to find the kinetic energy we first need to know the concept behind alpha particles.Alpha particles have a similarity to the nucleus of the Helium atom in which two protons and two neutrons are present while there are two electrons present in the outermost shell known as the valence shell of the atom.

We know that there is a tendency of atoms to exist in other oxidation states either the loss or gain of electrons to its valence shell. The same alpha particle or Helium atom can exist in various forms. In order to find the charge of the alpha particle we consider this.We know that the atom has a tendency to either lose electrons or gain electrons based on whether electrons are being lost, that is, removed from the valence shell or added to the valence shell. The charge of the electrons are said to be negative and hence when electrons leave the valence shell, that is, when electrons are lost the overall charge of the atoms becomes positive.

This is what happens with Helium atoms as well as alpha particles as well. When considering the alpha particles when two electrons are lost the alpha particle is said to attain an oxidation state of $ + 2$ state and thus by considering the ${\alpha ^{2 + }}$ state of the alpha particle we can say that the charge of the alpha particle is $ + 2e$ where $e$ stands for the charge of a single electron and the plus symbol indicates that the electrons are lost.
Thus, we can say that the charge $q$ of the alpha particle is:
$q = + 2e$ --------($1$)

Let us now extract the data from the given question. The potential difference which makes the alpha particles to accelerate is given as ${10^6}$ volt. We are asked to find out the kinetic energy of the alpha particle. Since, the question gives that the particle considered here is an alpha particle we have determined the charge of the alpha particle already. We know that the work done by the alpha particle in accelerating through a distance is given by the equation:
$W = qV$ -----------($2$)
Here, in accordance with the work-energy theorem, we consider the kinetic energy to be equivalent to the work done by the alpha particle in travelling through a displacement due to the potential difference that is applied. The potential difference that is applied in turn exerts a force on it which makes it accelerate.

We know that the work done is the amount of force required to move the particle by a distance and hence in order for the alpha particle to move or to get it in motion it needs to acquire some kinetic energy and hence this the work done in bringing it into its original state of rest the same amount of work must be done. The energy contained by the body in motion is the kinetic energy and hence this is equivalent to the work done on the alpha particle to make it accelerate.Thus, since the work done is equal to the kinetic energy, $K$ of the particle equation ($2$) becomes:
$K = qV$ --------($3$)
We are already given the potential difference and we have already determined the charge of the alpha particle.

Given, $V = {10^6}V$. Hence by substituting the above potential difference value and the charge of the alpha particle from equation ($1$) we get:
$K = \left( { + 2e} \right){10^6}V$
$ \Rightarrow K = 2 \times {10^6}eV$
Here, $eV$ indicating electron- volt will be the unit of kinetic energy. We know the conversion:
$1\,MeV = {10^6}\,eV$
This means that the kinetic energy in terms of mega electron volt can be determined from the above conversion relation. Hence, by unitary method we get:
$ \Rightarrow 2 \times {10^6}eV = 2MeV$
$ \therefore K = 2MeV$
Hence, the kinetic energy of the particle is equal to $2\,MeV$.

Therefore, the correct answer is option B.

Additional information:Since, alpha particles have an overall positive charge they tend to deflect towards negatively charged plates and beta particles tend to deflect towards positively charged plates when performing experiments and hence this is a way to determine or identify the radiation that is present.

Note: The alpha particles are actually heavier in comparison to beta particles which are the fast moving electrons that are emitted out of beta radiation. The common mistake that can be made in this problem can be the determining the charge of the alpha particle which must be taken as the Helium atom with an oxidation state of plus two charge instead of the normal alpha particle. The kinetic energy must be considered to be the work done of the alpha particle which is also often not considered which is also another common error.