Question

In Thomson experiment of finding $\frac{e}{m}$ for electrons, beam of electrons is replaced by that of muons (particle with same charge as of electrons but mass 208 times that of electrons).
No deflection condition in this is case satisfied, if :

• A. B is increased 208 times
• B. E is increased 208 times
• C. B is increased 14.4 times
• D. None of the above

Answer 1

The correct option is C B is increased 14.4 times

We have $qV=\frac{1}{2}m{v}^2$
$v=\sqrt{\frac{2qV}{m}}$
$qvB=qE\Rightarrow B=\frac{E}{V}$
$B=\frac{E}{\sqrt{\frac{2qV}{m}}}=\frac{E\sqrt{m}}{\sqrt{2qV}}$
$B_1=\frac{E\sqrt{m_e}}{\sqrt{2eV}}$ and $B_2=\frac{E\sqrt{m_{\mu }}}{\sqrt{2eV}}$
$\frac{B_2}{B_1}=\sqrt{\frac{m_{\mu }}{m_e}}$
$B_1=\sqrt{208}=14.4$
$B_2=14.4B_1$