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

  • Option 1)

    1 MeV

  • Option 2)

    0.5 MeV

  • Option 3)

    4 MeV

     

  • Option 4)

    2 MeV

 

Answers (1)

best_answer

Radius of a particle moving in magnetic field is given by

R =\frac{mv}{qB}= \sqrt{\frac{2.mE}{qB}}

E = Kinetic energy

Rproton = R\alpha-particle

=> \frac{\sqrt{2.m_{p}E_{p}}}{q_{p}.B}=\frac{\sqrt{2m_{\alpha }.E_{\alpha }}}{q_{\alpha }.B}

\therefore \frac{E_{\alpha }}{E_{p}}=(\frac{m_{p}}{m_{\alpha }}).(\frac{q_{\alpha }}{q_{p}})^{2}=\frac{1}{4}.4=1

therefore, E\alpha= Ep= 1 MeV


Option 1)

1 MeV

this is the correct option

Option 2)

0.5 MeV

this is the incorrect option

Option 3)

4 MeV

 

this is the incorrect option

Option 4)

2 MeV

this is the incorrect option

Posted by

Plabita

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