A proton of mass m and charge -e is moving in a circular orbit of a magnetic field with energy 1MeV- What should be the energy of - -particle mass - 4 m and charge - -2e- so that it can revolve in the path of same radius-

By using r=√(2mK)/qB; r → same, B → same ⇒ K∝q^2/m Hence K_α/K_p= ( q_α/q_p ) ^2×m_p/m_α= ( 2q_p/q_p )^2×m_p/4m_p=1⇒ K_α =K_p=1 MeV. ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Physics

A proton of m...

Question

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of $α$ -particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius?

1 MeV

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

4 MeV

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

2 MeV

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

0.5 MeV

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

Open in App

Solution

The correct option is A 1 MeV
By using $r = \frac{\sqrt{2 m K}}{q B}$ ; $r \to$ same, $B \to$ same $\Rightarrow K \propto \frac{q^{2}}{m}$
Hence $\frac{K_{α}}{K_{p}} = {(\frac{q_{α}}{q_{p}})}^{2} \times \frac{m_{p}}{m_{α}} = {(\frac{2 q_{p}}{q_{p}})}^{2} \times \frac{m_{p}}{4 m_{p}} = 1 \Rightarrow K_{α} = K_{p} = 1 M e V$ .

Suggest Corrections

Similar questions

Q. A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of

α

-particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius?

Q. A proton of a mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1 MeV. What should be the energy of a particle (mass 4m and charge +2e) so that it can revolve in the path of same radius?

Q. A proton carrying

1 M e V

kinetic energy is moving in a circular path of radius

R

in uniform magnetic field. What should be the energy of an

α

-particle to describe a circle of same radius in the same field?

A proton with 1 MeV kinetic energy is moving in a circular path of radius R in a uniform magnetic field. What should be the energy of an α - particle to describe a circle of same radius in the same magnetic field?

Q. A proton of mass

m

and charge

q

is moving in a plane with kinetic energy

E

. If there exists a uniform magnetic field

B

, perpendicular to the plane of the motion, the proton will move in a circular path of radius:

Join BYJU'S Learning Program

Explore more

Standard XI Physics

Join BYJU'S Learning Program

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of α -particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius?

The correct option is A 1 MeV​By using r=√2mKqB; r → same, B → same ⇒K∝q2m Hence KαKp=(qαqp)2×mpmα=(2qpqp)2×mp4mp=1⇒Kα=Kp=1MeV.

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of $α$ -particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius?