If the nucleus _ 13 ^ 27 Al has a nuclear radius of about 3.6 fm , then _ 32 ^ 125 Te would have its radius approximately as:

Correct answer: D. 6.0 fm . Nuclear radii (R= (R_0) A^ 1 / 3 ) where, A is the mass number. ( R_ Te R_ Al = ( A_ Te A_ Al ) ^ 1 / 3 = ((125)/(27)) ^ 1 / 3 =((5)/(3))) or, ( R_ Te =(5)/(3) _ Al =(5)/(3) ×3.6=6 fm). (Given (R_ Al =3.6 fm ))

NEET 2007

Exam: NEET
Year: 2007
Subject: Physics
Chapter: Nuclear Physics

NEET 2007 · Physics · Nuclear Physics

If the nucleus _ 13 ^ 27 Al has a nuclear radius of about 3.6 fm ,...

If the nucleus

{ }_{13}^{27} \mathrm{Al}

has a nuclear radius of about

3.6 \mathrm{fm}

, then

{ }_{32}^{125} \mathrm{Te}

would have its radius approximately as:

Options

A $9.6 \mathrm{fm}$
B $12.0 \mathrm{fm}$
C $4.8 \mathrm{fm}$
D $6.0 \mathrm{fm}$ .

Correct answer

D. $6.0 \mathrm{fm}$ .

Step-by-step solution

Nuclear radii (R=

\left(R_0\right)

A^{1 / 3}) where,

A

is the mass number. (\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=

\left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)

^{1 / 3}= \left(\frac{125}{27}\right)

^{1 / 3}=\left(\frac{5}{3}\right)) or, (\quad R_{\mathrm{Te}}=\frac{5}{3} \timesR_{\mathrm{Al}}=\frac{5}{3} \times3.6=6 \mathrm{fm}). (Given (R_{\mathrm{Al}}=3.6 \mathrm{fm}

))

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