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An $\alpha -$ particle of $1\,MeV$ energy moves on circular path in uniform magnetic field. Then kinetic energy of proton in same magnetic field for circular path of double radius is......$MeV$
$1$
$4$
$2$
None
Solution
${{\text{E}}_{\text{K}}} = \frac{{{{\text{q}}^2}{{\text{B}}^2}{{\text{r}}^2}}}{{2{\text{m}}}} \propto \frac{{{{\text{q}}^2}{{\text{r}}^2}}}{{\text{m}}}$
$\frac{{{{\left( {{{\text{E}}_{\text{K}}}} \right)}_{\text{p}}}}}{{{{\left( {{{\text{E}}_{\text{K}}}} \right)}_\alpha }}} = {\left( {\frac{{{{\text{q}}^2}{{\text{r}}^2}}}{{\text{m}}}} \right)_{\text{p}}} \times {\left( {\frac{{\text{m}}}{{{{\text{q}}^2}{{\text{r}}^2}}}} \right)_\alpha }$
$ = \frac{{{e^2}{{(2{\text{r}})}^2}}}{{{{\text{m}}_{\text{p}}}}} \times \frac{{4{{\text{m}}_{\text{p}}}}}{{{{(2e)}^2}{{\text{r}}^2}}}$
$ = \frac{{4{{\text{e}}^2}{{\text{r}}^2}}}{{{{\text{m}}_{\text{p}}}}} \times \frac{{4{{\text{m}}_{\text{p}}}}}{{4{{\text{e}}^2}{{\text{r}}^2}}} = 4$
${\left( {{{\text{E}}_{\text{K}}}} \right)_{\text{p}}} = 4{\left( {{{\text{E}}_{\text{K}}}} \right)_\alpha } = 4\,{\text{MeV}}$
