Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A proton and a deuteron initially at rest, are accelerated through the same potential difference. Which of the following statements or statements about the final properties of the two particles is true?
A) They have different speeds.
B) They have the same momentum.
C) They have the same kinetic energy.
D) None of these.

seo-qna
SearchIcon
Answer
VerifiedVerified
103.5k+ views
Hint: Deuteron consists of a proton and a neutron and will thus have a mass greater than the proton. The conservation of energy theorem suggests that the kinetic energy of the proton or deuteron must be equal to their respective potential energy.

Formula Used:
1) The potential energy of a particle placed in a potential difference $\Delta V$ is given by, $U = q\Delta V$ where $q$ is the charge of the particle.
2) The kinetic energy of a particle of mass $m$ moving with velocity $v$ is given by, $K = \dfrac{1}{2}m{v^2}$ .
3) The momentum of a particle of kinetic energy $K$ can be expressed as $p = \sqrt {2mK} $ where $m$ is the mass of the particle.

Complete step by step answer:
Step 1: List the key features of proton and deuteron.
A deuteron has one neutron and one proton.
The charge of the proton is ${q_p} = + e$ and the charge of the deuteron is ${q_D} = + e$ .
The mass of the deuteron is, however, greater than the mass of the proton i.e., ${m_D} > {m_p}$ .
Both of the particles are initially at rest and are then accelerated through the same potential difference $\Delta V$ .
Step 2: Find the kinetic energies of the two particles.
The total energy of the particle (proton and deuteron) must be conserved. Since the two particles are initially at rest, they do not possess kinetic energy before getting accelerated but posses potential energy.
So, by applying the conservation of energy we can say that the kinetic energy of the particle (proton or deuteron) must be equal to the potential energy of the particle (proton or deuteron).
The potential energy of a particle placed in a potential difference $\Delta V$ is given by, $U = q\Delta V$ where $q$ is the charge of the particle.
For proton having a charge ${q_p} = + e$, the potential energy will be ${U_p} = e\Delta V$ and thus its kinetic energy will also be ${K_p} = e\Delta V$ ------ (1)
For deuteron having a charge ${q_D} = + e$, the potential energy will be ${U_D} = e\Delta V$ and thus its kinetic energy will also be ${K_D} = e\Delta V$ ------ (2)
From equation (1) and (2) we can conclude that the kinetic energy of the proton and that of the deuteron are equal.
Step 3: Check to see if they have the same speed.
The kinetic energy of the proton of mass ${m_p}$ moving with velocity ${v_p}$ will be
${K_p} = \dfrac{1}{2}{m_p}v_p^2$ --------- (4)
Similarly, the kinetic energy of deuteron of mass ${m_D}$ moving with velocity ${v_D}$ will be
${K_D} = \dfrac{1}{2}{m_D}v_D^2$ --------- (5)
Since we have established that ${K_p} = {K_D} = K$, we can equate (4) and (5), i.e., $\dfrac{1}{2}{m_p}v_p^2 = \dfrac{1}{2}{m_D}v_D^2$
This can be expressed as the ratio of their masses as $\dfrac{{{m_p}}}{{{m_D}}} = \dfrac{{v_D^2}}{{v_p^2}}$ ------ (6)
Since we know ${m_D} > {m_p}$, equation (6) will imply that ${v_D} < {v_p}$ .
So, they have different speeds.
Step 4: Check to see if they have the same momentum.
The momentum of protons of kinetic energy $K$ and mass ${m_p}$ can be expressed as ${p_p} = \sqrt {2{m_p}K} $ .
Similarly, the momentum of a deuteron of kinetic energy $K$ and mass ${m_D}$ can be expressed as ${p_D} = \sqrt {2{m_D}K} $ .
Since ${m_D} > {m_p}$, ${p_p} \ne {p_D}$
Thus they do not have the same momentum.
So, the correct options are A and C.

Note: Neutrons are chargeless particles and so the presence of one neutron in a deuteron will not affect its total charge. Here, there is no additional work done or heat transferred. So, the total energy must be constant. The total energy is the sum of the particle’s potential energy and its kinetic energy. It is the loss of the potential energy of the proton (or deuteron) that turns into an increase in its kinetic energy.