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PhysicsModern PhysicsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

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

Options:

Answer:

Solution:

$r = \frac{\sqrt{2 m K}}{q B} \Rightarrow K \propto \frac{q^{2}}{m} \Rightarrow \frac{K_{p}}{K_{α}} = {(\frac{q_{p}}{q_{α}})}^{2} \times \frac{m_{α}}{m_{p}}$

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

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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