Answer
$$
3.2 \mathrm{\ kg}
$$
Work Step by Step
$\text { The amount of }^{90}$ Sr needed is
$$
M=\frac{150 \mathrm{W}}{(0.050)(0.93 \mathrm{W} / \mathrm{g})}=3.2 \mathrm{\ kg}
$$
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 1331: 24b
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.
