Answer
$K_\alpha=3.541 \mathrm{MeV}$
Work Step by Step
$
K_\alpha=\frac{Q}{1+\left(m_\alpha / m_{\mathrm{n}}\right)}\\=\frac{17.59 \mathrm{MeV}}{1+(4.0015 \mathrm{u} / 1.008665 \mathrm{u})}\\=3.541 \mathrm{MeV}
$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 43 - Energy from the Nucleus - Problems - Page 1333: 52a
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