QJEEM18Pha915: JEE Main 2018 Physics - Class 12 JEEM18Pha915 Solution: An electron, a proton and an alpha particle having the same…

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 _$\propto$