JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 18)
A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths $${{{\lambda _A}} \over {{\lambda _B}}}$$ is close to :
4.47
10.00
14.14
0.07
Explanation
K.E. acquired by charge = K = qV
$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2mK} }}$$ = $${h \over {\sqrt {2mqV} }}$$
$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50} $$
$$ = 2 \times 7.07 = 14.14$$
$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2mK} }}$$ = $${h \over {\sqrt {2mqV} }}$$
$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50} $$
$$ = 2 \times 7.07 = 14.14$$
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