College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 7 - Problems - Page 263: 26


The recoil speed of the thorium nucleus is $~2.57\times 10^5~m/s$

Work Step by Step

The initial momentum of the system is zero. By conservation of momentum, if the alpha particle gains a certain momentum $p_a$ in one direction, the remaining nucleus will gain the same magnitude of momentum $p_t$ in the opposite direction. We can find the recoil speed $v_r$ of the thorium nucleus: $p_t = p_a$ $m_t~v_r = m_a~v_a$ $v_r = \frac{m_a~v_a}{m_s}$ $v_r = \frac{(4.0~u)(0.050~c)}{234~u}$ $v_r = 8.55\times 10^{-4}~c$ $v_r = (8.55\times 10^{-4})(3.0\times 10^8~m/s)$ $v_r = 2.57\times 10^5~m/s$ The recoil speed of the thorium nucleus is $~2.57\times 10^5~m/s$.
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