Atoms And Nuclei Question 333
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:
- 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
