A proton, an electron and a Helium nucleus, have the same energy.

They are in circular orbits in a plane due to magnetic field perpendicular 

to the plane. Let r_{p}, r_{e} \: \: and\: \: r_{He} be their respective radii, then,

  • Option 1)

    r_{e}>r_{p} = r_{He}

  • Option 2)

    r_{e}<r_{p} = r_{He}

  • Option 3)

    r_{e}<r_{p} < r_{He}

  • Option 4)

    r_{e}>r_{p} > r_{He}

 

Answers (1)

 

Radius of charged particle -

r=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{2mK}}{qB}=\frac{1}{B}\sqrt{}\frac{2mV}{q}

m=mass of particle 

P=momentum of particle 

k= kinetic energy of particle 

-

 

 r=\frac{mv}{qB}=\frac{\sqrt{2mk}}{qB}

=> r\propto \frac{\sqrt m}{q}

               proton      electron      He^{+} 

m             m_p            m_e           4m_p

q              e                 e                 e

and    m_p>m_c

So,

r_He^{+}>r_p>r_e


Option 1)

r_{e}>r_{p} = r_{He}

Option 2)

r_{e}<r_{p} = r_{He}

Option 3)

r_{e}<r_{p} < r_{He}

Option 4)

r_{e}>r_{p} > r_{He}

Exams
Articles
Questions