Question

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 25.0° above its initial direction of motion, and the oxygen nucleus recoils at an angle of 50.0° below this initial direction. The final speed of the oxygen nucleus is 2.08×105 m/s. (The mass of an alpha particle is 4.0 u, and the mass of an oxygen nucleus is 16 u.) What is the final speed of the alpha particle?

Answer 1

Answer:

$v_{i}= 19\times 10^5\ m/s$

Explanation:

given,

scattering angle of alpha particle = 25.0°  above its initial direction of motion

oxygen nucleus recoils at = 50.0° below this initial direction.

final speed of the oxygen = 2.08×10⁵ m/s

mass of alpha particle = 4.0 u

mass o oxygen nucleus = 16 u

momentum conservation along x- axis

$m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta$

$4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0$

$v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}$....(1)

Along y-direction

$0 = m_av_a sin \theta - m_ov_o sin\theta$

$0 = 4\times v_a sin 25 - 16\times 2.08 \times 10^5 sin50^0$

$v_a = \dfrac{25.49 \times 10^5}{1.69}$

$v_a = 15.082\times 10^5\ m/s$

putting value in equation (1)

$v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}$

$v_{i}= 19\times 10^5\ m/s$