JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Magnetic Effects of Current

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A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let

r_{p}, r_{e}

and

r_{H e}

be their respective radii, then,

(A) $r_{e} > r_{p} = r_{He}$
(B) $r_{e} > r_{p} > r_{He}$
(C) $r_{e} < r_{p} < r_{He}$
(D) $r_{e} < r_{p} = r_{He}$ ✓ Correct

Answer: (D)

Solution

For a charged particle bent by a magnetic field,

r = \frac{m v}{q B} = \frac{2 m K}{q B}

, so at a common kinetic energy

K

the radius scales as

\frac{m}{q}

. For the proton this is

\frac{m _{p}}{e}

, for the electron

\frac{m _{e}}{e}

, and for the helium nucleus

\frac{4 m _{p}}{2 e} = \frac{m _{p}}{e}

. Since the electron is far lighter than the proton its radius is the smallest, while the proton and the helium nucleus share the same radius:

r_{e} < r_{p} = r_{H e}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Magnetic Effects of Current

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