JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 27)
If the de Broglie wavelengths associated with a proton and an $$\alpha $$-particle are equal, then the ratio of velocities of the proton and the $$\alpha $$-particle will be :
4 : 1
2 : 1
1 : 2
1 : 4
Explanation
We know, $${\lambda _p} = {h \over {{p_p}}}$$ = $$ {h \over {{m_p}{v_p}}}$$
Similarly, $${\lambda _\alpha } = {h \over {{m_\alpha }{v_\alpha }}}$$
Given, $${\lambda _p} = {\lambda _\alpha }$$
$$ \Rightarrow $$ $${h \over {{m_p}{v_p}}} = {h \over {{m_\alpha }{v_\alpha }}}$$
$$ \therefore $$ $${{{v_p}} \over {{v_\alpha }}} = {{{m_\alpha }} \over {{m_p}}}$$ = $${{4{m_p}} \over {{m_p}}} = {4 \over 1}$$
Similarly, $${\lambda _\alpha } = {h \over {{m_\alpha }{v_\alpha }}}$$
Given, $${\lambda _p} = {\lambda _\alpha }$$
$$ \Rightarrow $$ $${h \over {{m_p}{v_p}}} = {h \over {{m_\alpha }{v_\alpha }}}$$
$$ \therefore $$ $${{{v_p}} \over {{v_\alpha }}} = {{{m_\alpha }} \over {{m_p}}}$$ = $${{4{m_p}} \over {{m_p}}} = {4 \over 1}$$
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