Answer:
option B
Explanation:
we know,
change in energy is equal to
[tex]W = \dfrac{1}{2}m(v^2 - u^2)[/tex]
[tex]W = \dfrac{1}{2}m(v^2 - 0^2)[/tex]
[tex]W = \dfrac{1}{2}m v^2[/tex]
[tex]q = \dfrac{1}{2}m v^2[/tex]
proton mass and the neutron mass are roughly the same
so,
[tex]q \alpha m v^2[/tex]
now,
[tex]\dfrac{q_p}{q_{\alpha}} = \dfrac{m_p v_p^2}{m_{\alpha}v_{\alpha}^2}[/tex]
[tex]\dfrac{q_p}{q_{\alpha}} = \dfrac{m_p v_p^2}{2 m_pv_{\alpha}^2}[/tex]
we know,
mass of alpha particle is four times mass of the mass of proton.
mα = 4 m_p
[tex]\dfrac{e}{2e} = \dfrac{ v_p^2}{4 v_{\alpha}^2}[/tex]
[tex] \dfrac{ v_p^2}{v_{\alpha}^2} = 2[/tex]
[tex]v_{\alpha}^2 =\dfrac{ v_p^2}{2}[/tex]
[tex]v_{\alpha}=\dfrac{ v_p}{\sqrt{2}}[/tex]
less by a factor of √2
Hence, the correct answer is option B
