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. ...

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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

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4 MeV

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2 MeV

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0.5 MeV

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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 $\frac{K_{α}}{K_{p}} = {(\frac{q_{α}}{q_{p}})}^{2} \times \frac{m_{p}}{m_{α}} = {(\frac{2 q_{p}}{q_{p}})}^{2} \times \frac{m_{p}}{4 m_{p}} = 1 \Rightarrow K_{α} = K_{p} = 1 M e V$ .

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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?