When a charged particle moves in a magnetic field then the charged particle moves in a circular path. So,

$${{m{v^2}} \over r}$$ = Bqv

$$ \Rightarrow $$$$\,\,\,$$ r = $${{mv} \over {Bq}}$$

We know kinetic energy, K = $${1 \over 2}$$ mv²

$$\therefore\,\,\,$$ mv = $$\sqrt {2Km} $$

$$\therefore\,\,\,$$ r = $${{\sqrt {2Km} } \over {Bq}}$$

According to the question,

K_e (electron) = K_p (proton) = K_{$$\alpha $$}(Alpha particle) = K = constant, and all of them are in uniform magnetic field.

$$ \therefore $$ B = constant.

$$\therefore\,\,\,$$ r $$ \propto $$ $${{\sqrt m } \over q}$$

For proton (₁H¹), mass = m, and charge = e

$$\therefore\,\,\,$$ r_p $$ \propto $$ $${{\sqrt m } \over e}$$

For alpha particle (₂H⁴),

mass = 4m

and charge = 2e

$$\therefore\,\,\,$$ r_{$$ \alpha $$} $$ \propto $$ $${{\sqrt {4m} } \over {2e}}$$ $$ \propto $$ $${{\sqrt m } \over e}$$

$$ \therefore $$$$\,\,\,$$ r_p = r_{$$ \alpha$$}

For electron,

charge = e

and mass (m_e) = 9.1 $$ \times $$ 10^$$-$$31 kg

and mass of proton = 1.67 $$ \times $$ 10^$$-$$27 kg

$$\therefore\,\,\,$$ mass of electron < mass of proton.

r_e $$ \propto $$ $${{\sqrt {{m_e}} } \over e}$$ < r_p

$$\therefore\,\,\,$$ r_e < r_p = r_{$$ \propto $$}