Nuclear Physics And Radioactivity Question 403
Question: If the nucleus $ _{13}^{27}Al $ has nuclear radius of about 3.6 fm, then $ _{32}^{125}Te $ would have its radius approximately as
Options:
A) 9.6 fm
B) 12.0 fm
C) 4.8 fm
D) 6.0 fm.
Show Answer
Answer:
Correct Answer: D
Solution:
[d] It has been known that a nucleus of mass number A has radius $ R=R_{0}{{A}^{1/3}}, $ where $ R_{0}=1.2\times {{10}^{-15}}m $ and A = mass number In case of $ {13}^{27}A\ell , $ let nuclear radius be $ R{1} $ and for $ {32}^{125}Te, $ nuclear radius be $ R{2} $ For $ {13}^{27}Al,,R{1}=R_{0}{{(27)}^{1/3}}=3R_{0} $ For $ {32}^{125}Te,R{2}=R_{0}{{(125)}^{1/3}}=5R_{0} $ $ \frac{R_{2}}{R_{1}}=\frac{5R_{0}}{3R_{0}}=\frac{5}{3}R_{1}=\frac{5}{3}\times 3.6=6fm $
BETA
