1. A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is ⇒ (NEET 2022 Phase 1)

A. 1 : 1

B. 4 : 5

C. 5 : 4

D. 25 : 16

The correct answer is option (C)

Radius of nuclei with mass number A varies as

$R = R_{0} A^{1 / 3}$

$\frac{R_{1}}{R_{2}} = {(\frac{125}{64})}^{1 / 3} = \frac{5}{4} = 5 : 4$

2. If radius of the $_{13}^{27}$ Al nucleus is taken to be R_Al, then
the radius of $_{53}^{125}$ Te nucleus is nearly ⇒ (AIPMT 2015 Cancelled Paper)

A. $\frac{3}{5} R_{A l}$

B. ${(\frac{13}{53})}^{1 / 3} R_{A l}$

C. ${(\frac{53}{13})}^{1 / 3} R_{A l}$

D. $\frac{5}{3} R_{A l}$

The Correct Answer is Option (D)

As we know, R = R₀ (A)^1/3

where A = mass number

R_Al = R₀ (27)^1/3 = 3R₀

R_Te = R₀ (125)^1/3 = 5R₀ = $\frac{5}{3} R_{A l}$

3. If the nuclear radius of ²⁷Al is 3.6 fermi, the approximate nuclear radius of ⁶⁴Cu in fermi is ⇒ ( AIPMT 2012 Prelims)

A. 2.4

B. 1.2

C. 4.8

D. 3.6

The Correct Answer is Option (C)

Nuclear radius, R = R₀A^1/3

where R₀ is a constant and A is the mass number

$∴$ $\frac{R_{A l}}{R_{C u}} = \frac{{(27)}^{1 / 3}}{{(64)}^{1 / 3}}$ = $\frac{3}{4}$

$\Rightarrow$ R_Cu = $\frac{4}{3} \times R_{A l}$ = $\frac{4}{3} \times 3.6$ = 4.8 fermi

4.Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be ⇒ (AIPMT 2008)

A. ${(3)}^{1 / 3} : 1$

B. 1 : 1

C. 1 : 3

D. 3 : 1

The Correct Answer is Option (B)

A₁ : A₂ = 1 : 3

Their radii will be in the ratio

$R_{0} A_{1}^{1 / 3}$ : $R_{0} A_{2}^{1 / 3}$ = 1 : 3^1/3

$∴$ $ρ_{A_{1}}$ : $ρ_{A_{2}}$ = $\frac{A_{1}}{\frac{4}{3} π R_{1}^{3}}$ : $\frac{A_{2}}{\frac{4}{3} π R_{2}^{3}}$

= $\frac{1}{1^{3}} : \frac{3}{{(3^{1 / 3})}^{3}}$

= 1 : 1

5. If the nucleus $_{13}^{27} A l$ has a nuclear radius of about 3.6 fm, them $_{32}^{125} T e$ would have its radius approximately as ⇒ (AIPMT 2007)

A. 9.6 fm

B. 12.0 fm

C. 4.8 fm

D. 6.0 fm

The Correct Answer is Option (D)

A nucleus of mass number A has radius

R = R₀ A^1/3,

where R₀ = 1.2 × 10^–15 m and A = mass number

For $_{13}^{27} A l$ , R₁ = R₀ (27)^1/3 = 3R₀

For $_{32}^{125} T e$ , R₂ = R₀ (125)^1/3 = 5R₀

$∴$ $\frac{R_{2}}{R_{1}} = \frac{5 R_{0}}{3 R_{0}}$

$\Rightarrow$ $R_{2} = \frac{5}{3} \times R_{1}$ = $\frac{5}{3} \times 3.6$ = 6.0 fm

6. The radius of germanium (Ge) nuclide is measured to be twice the radius of $_{4}^{9}$ Be. The number of nucleons in Ge are ⇒ ( AIPMT 2006)

A. 72

B. 73

C. 74

D. 75

The Correct Answer is Option (A)

For berillium, R₁ = R₀ (9)^1/3

For germanium, R₂ = R₀ A^1/3

$\frac{R_{1}}{R_{2}}$ = $\frac{{(9)}^{1 / 3}}{A^{1 / 3}}$

$\Rightarrow$ $\frac{1}{2}$ = $\frac{{(9)}^{1 / 3}}{A^{1 / 3}}$

$\Rightarrow$ $\frac{1}{8} = \frac{9}{A}$

$\Rightarrow$ A = 72

7. Which one of the following pairs of nuclei are isotones ? ⇒ ( AIPMT 2005)

A. ₃₄Se⁷⁴, ₃₁Ga⁷¹

B. ₃₈Sr⁸⁴, ₃₈Sr⁸⁶

C. ₄₂Mo⁹², ₄₀Zr⁹²

D. ₂₀Ca⁴⁰, ₁₆S³²

The Correct Answer is Option (A)

Isotones means number of neutron remains same.

8. A nucleus represented by the symbol $_{Z}^{A} X$ has ⇒ (AIPMT 2004)

A. Z neutrons and A $-$ Z protons

B. Z protons and A $-$ Z neutrons

C. Z protons and A neutrons

D. A protons and Z $-$ A neutrons

The Correct Answer is Option (B)

Z is number of protons and A is the total number of protons and neutrons.

9. The volume occupied by an atom is greater than the volume of the nucleus by a factor of about ⇒ (AIPMT 2003)

A. 10¹

B. 10⁵

C. 10¹⁰

D. 10¹⁵

The Correct Answer is Option (D)

As V $\propto$ r³

$\Rightarrow$ V_atom = $\frac{r_{a t o m}^{3}}{r_{n u c l e u s}^{3}}$ $\times$ V_nucleus

= $\frac{{(10^{- 10})}^{3}}{{(10^{- 15})}^{3}}$ $\times$ V_nucleus

= 10¹⁵ $\times$ V_nucleus

10. The mass number of a nucleus is ⇒ (AIPMT 2003)

A. always less than its atomic number

B. always more than its atomic number

C. sometimes equal to its atomic number

D. sometimes less than and sometimes more than its atomic number

The Correct Answer is Option (C)

Mass number = atomic number + no. of neutrons

For hydrogen, number of neutrons = 0

So, mass number = Atomic number.

Hence mass number is sometimes equal to atomic number.

11. For the given reaction, the particle X is
$_{6} C^{11} \to_{5} B^{11} + β^{+} + X$ ⇒ ( AIPMT 2000)

A. neutron

B. anti neutrino

C. neutrino

D. proton

The Correct Answer is Option (C)

The particle X is neutrino.

⇒ Nuclei

Topic 1 : Composition and Size of Nucleus