Question

A proton and an alpha particle (helium nucleus consisting of two protons and two neutrons) are accelerated from rest across the same potential difference. Assume the proton mass and the neutron mass are roughly the same and neglect any relativistic effect. Compared to the final speed of the proton, the final speed of the alpha particle is?1. less by a factor of 22. less by a factor of √ 23. less by a factor of 44. greater by a factor of 25. the same

Answer 1

option B

Step-by-step explanation:

we know,

change in energy is equal to

`W = (1)/(2)m(v^2 - u^2)`

`W = (1)/(2)m(v^2 - 0^2)`

`W = (1)/(2)m v^2`

`q = (1)/(2)m v^2`

proton mass and the neutron mass are roughly the same

so,

`q \alpha m v^2`

now,

`(q_p)/(q_(\alpha)) = (m_p v_p^2)/(m_(\alpha)v_(\alpha)^2)`

`(q_p)/(q_(\alpha)) = (m_p v_p^2)/(2 m_pv_(\alpha)^2)`

we know,

mass of alpha particle is four times mass of the mass of proton.

mα = 4 m_p

`(e)/(2e) = ( v_p^2)/(4 v_(\alpha)^2)`

`( v_p^2)/(v_(\alpha)^2) = 2`

`v_(\alpha)^2 =( v_p^2)/(2)`

`v_(\alpha)=( v_p)/(√(2))`

less by a factor of √2

Hence, the correct answer is option B