CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A proton of energy 8eV is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be


A

4eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

2eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

8eV

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

6eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

8eV


Explanation for the correct option:

Step 1: Given

Energy of proton, Kp=8eV

Let B be the magnetic field and r be the radius of the circular path.
It is given that these are the same for both the proton and the α-particle .

Step 2: Formulas used

We know that, the momentum of a particle moving in a magnetic field B is, p=mv=qBr
where m is the mass of the particle, v is its velocity, q is its charge, and r is the radius of its circular path.

We also know that kinetic energy, K=12mv2

Thus,

p2=m2v2mv2=p2mK=12p2mK=q2B2r22m

Step 3: Calculate the energy of α-particle

The kinetic energy of the proton can be written as,
Kp=qp2B2r22mp

An α-particle is simply the nucleus of a helium-4 (He24) atom.
Thus the mass of the α-particle is 4 times that of the proton i.e., mα=4mp (mα is the mass of the α-particle and mp is the mass of the proton)

The α-particle has two protons thus its charge is twice that of a proton i.e., qα=2qp (qα is the charge of the α-particle and qp is that of the proton).

Thus,

Kα=2qp2B2r224mpKα=4qp2B2r22×4mpKα=qp2B2r22mpKα=KpKα=8eV

Hence, option C is correct.


flag
Suggest Corrections
thumbs-up
1
BNAT
mid-banner-image