JEE PYQ: Atomic Physics Question 51
Question 51 - 2019 (10 Apr Shift 1)
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,
(1) $r_e > r_p = r_{He}$
(2) $r_e < r_p = r_{He}$
(3) $r_e < r_p < r_{He}$
(4) $r_e > r_p > r_{He}$
Show Answer
Answer: (2)
Solution
As $mvr = qvB \Rightarrow r = \frac{mv}{qB} = \frac{\sqrt{2mK.E.}}{qB}$. For proton, electron and $\alpha$-particle, $m_{He} = 4m_p$ and $m_p » m_e$. Also $q_{He} = 2q_p$ and $q_e = q_p$. As KE of all the particles is same, $r \propto \frac{\sqrt{m}}{q}$. So $r_{He} = r_p > r_e$.
BETA
