Question

An atom of beryllium (m 8.00 u) splits into two atoms of helium (m-4.00 u) with the release of 92.2 keV of energy. Suppose the beryllium atom moves in the positive x direction and has kinetic energy 59.9 keV. One helium atom is found to move at an angle of 31° with respect to the x axis. Find the direction of motion of the second helium atom and find the velocity of each helium atom Work this problem in the following two ways. (Hint: Consider one helium to be eitted with velocity components Vx and Vy in the beryllium rest frame. What is the reship between Vx and vy? How do Vx and vy change when we move in the x direction at speed v? Assume the angle for the first helium atom is measured counterclockwise from the +x axis.) (a) by direct application of conservation of momentum and energy velocity af the first helium (magnitude) velocity of the second helium (magnitude) direction of motion of the second helium m/s m/s clockwise from the -x axis (b) by applying the results of a modified problem (the original beryllium atom is at rest) to a frame of reference moving with the original beryllium atom and then switching to the reference frame in which the beryllium is moving velocity of the first helium (magnitude) velocity of the second helium (magnitude) direction of motion of the second helium m/s m/s o clackwise from the -x axis

Answer 1

Given that m_be = 8.00 mu m_he = 4.00 mu Energy released = 92.2 kev K_be = 59.9 kev (a) As man of helium aton is same Apply conversation of linear momentum m_be mu = m_He V + m_He v^1 8mu = 4mv + 4mv^1 2mu = v + v^1 K.E_be = 1/2 m_be u^2 u squareroot 2K.E_be/m_be u = squareroot 2 times 59.9 times 10^3 times 1.6 times 10^-19/84 times (1.67 times 10^-27/4) = squareroot 191.68 times 10^-16/13.36 times 10^-27 \r\n = squareroot 14.347 times 10^11 u = 11.978 times^5 u = 1.1978 times 10^6 m/s From conservation of energy K E_be + K E_release = 1/2 mHe V^2 + 1/2m_He V_1 = 1/2 m_He (V^2 + (2u-u)^2) (59.9 + 92.2) times 10^3 times 1.6 times 10^19 = 1/2 times 4 times 1.67 times 10^-27[2v^2 + 4u^2 - 4uv] 152.1 times 10^-16 times 1.6 = 6.68/2 times 10^-27[2v^2 + 4u^2 - 4uv] 2.43 times^-14 = 3.34 times 10^-24 [2v^2 + 4u^2 4uv] 2.43 times 10^-14/3.34 times 10^-24 = 2v^2 + 4 times(1.02 times 10^6)^2 - 4 times 1.02 times 10^6v 0.727 times 10^10 = 2v^2 + 4 times 1.04 times 10^2 - 4.08 times 10^6v 0.727 times 10^10 -4.16 times 10^12 = 2v^2 - 4.08 times 10^6v 3.433 times 10^12 = 2v^2 - 4.08 times 10^6v 2v^2 - 4.08 times 10^6v - 3.433 times 10^12 = 0 v^2 - 2.04 times 10^6v - 1.717 times 10^12 = 0 \r\n solving it gives using quadratic formula v = (1.012, - 3.789)