Answer
$Q_1 = 5.98 MeV$
Work Step by Step
For the second reaction,
$^{223}Ra \longrightarrow ^{219}Rn + ^{4}He $
$Q_1 = (m_{Ra} - m_{Rn} - m_{He}) c^2$
$Q_1 = (223.01850u - 219.00948u - 4.00260 u) (931.5 MeV/u) $
$Q_1 = 5.98 MeV$
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 42 - Nuclear Physics - Problems - Page 1305: 52b
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
