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ATOMS

145263 If the ratio of radius of a nucleus with 61 neutrons to that of helium nucleus is 3 , then the atomic number of this nucleus is

1 27

2 47

3 51

4 61

5 108

Explanation:

B Given that,
Ratio of the radii $(\frac{R_{1}}{R_{2}}) = 3$
Number of neutrons $= (A - Z) = 61$
Where, $A =$ Atomic mass
$Z =$ Atomic number
We know that, radius of the nucleus $R = R_{0} (A)^{1 / 3}$
i.e., $R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$(3)^{3} = \frac{A_{1}}{A_{2}} \Rightarrow 27 = \frac{A_{1}}{A_{2}}$
$A_{1} = 27 \times A_{2}$
$A_{1} = 27 \times 4 {A_{2} =$ mass of helium $}$
$A_{1} = 108$
$∵ A - Z = 61$
$Z = A - 61 = 108 - 61 {∵ A_{1} = A}$
$Z = 47$

Kerala CEE - 2016

ATOMS

145264 The ratio of the surface area of the nuclei $_{52} {Te}^{125}$ to that of $_{13} {Al}^{27}$ is

1

\frac{5}{3}

2

\frac{125}{17}

3

\frac{1}{4}

4

\frac{25}{9}

5

\frac{3}{5}

Explanation:

D Given,
Atomic mass of $_{52} {Te}^{125} (A) = 125$
Atomic mass of $_{13} A l^{27} (A) = 27$
The radius of the nucleus
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
$R \propto (A)^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = {(\frac{125}{27})}^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$∵$ Surface area $A \propto R^{2}$
$∴ \frac{R_{Te}}{R_{Al}} = {(\frac{R_{Te}}{R_{Al}})}^{2} = \frac{R_{Te}}{R_{Al}} = {(\frac{5}{3})}^{2} = \frac{25}{9}$

Kerala CEE- 2014

ATOMS

145265 The ratio between the radii of nuclei with mass numbers 27 and 125 is

1

5 : 3

2

3 : 5

3

27 : 125

4

125 : 27

5

1 : 1

Explanation:

B Given,
$Mass (A_{1}) = 27$
$Mass (A_{2}) = 125$
We know that, radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant
$R =$ radius of nucleus
$R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{27}{125})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = \frac{3}{5}$
Hence, $R_{1} : R_{2} = 3 : 5$

Kerala CEE 2012

ATOMS

145266 The nuclear radius of a certain nucleus is $7.2$
$fm$ and it has charge of $1.28 \times 10^{- 17} C$ . The number of neutrons inside the nucleus is

1 136

2 142

3 140

4 132

5 126

Explanation:

A Given,
Radius of nucleus $(R) = 7.2 fm$
Charge of the nucleus $= 1.28 \times 10^{- 17} C$
We know that,
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} = 1.2 fm$
$A = {(\frac{R}{R_{0}})}^{3}$
$A = {(\frac{7.2}{1.2})}^{3} = (6)^{3} = 216$
Number of proton in the nucleus,
$Z = \frac{Charge of the nucleus}{Charge of a proton}$
$Z = \frac{1.28 \times 10^{- 17}}{1.6 \times 10^{- 19}}$
$Z = 80$
The number of neutrons in the nucleus $= A - Z$
$= 216 - 80$
$= 136$

Kerala CEE - 2008

ATOMS

145267 Radius of $_{2} {He}^{4}$ nucleus is 3 fermi. The radius of $_{82} {Pb}^{206}$ nucleus will be

1 5 fermi

2 6 fermi

3 11.16 fermi

4 8 fermi

Explanation:

C Given,
Radius of $_{2} {He}^{4}$ nucleus $(R_{1}) = 3 fm$
Radius of $_{82} {Pb}^{206}$ nucleus $(R_{2}) =$ ?
Mass of $_{2} {He}^{4} (A_{1}) = 4$
Mass of $_{82} {Pb}^{206} (A_{2}) = 206$
We know that,
$R = R_{0} (A)^{1 / 3}$
Here, $R \propto A^{1 / 3}$
$\frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3}$
$\frac{R_{2}}{3} = {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 3 \times {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 11.16 fm$

JCECE-2010

ATOMS

145263 If the ratio of radius of a nucleus with 61 neutrons to that of helium nucleus is 3 , then the atomic number of this nucleus is

1 27

2 47

3 51

4 61

5 108

Explanation:

B Given that,
Ratio of the radii $(\frac{R_{1}}{R_{2}}) = 3$
Number of neutrons $= (A - Z) = 61$
Where, $A =$ Atomic mass
$Z =$ Atomic number
We know that, radius of the nucleus $R = R_{0} (A)^{1 / 3}$
i.e., $R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$(3)^{3} = \frac{A_{1}}{A_{2}} \Rightarrow 27 = \frac{A_{1}}{A_{2}}$
$A_{1} = 27 \times A_{2}$
$A_{1} = 27 \times 4 {A_{2} =$ mass of helium $}$
$A_{1} = 108$
$∵ A - Z = 61$
$Z = A - 61 = 108 - 61 {∵ A_{1} = A}$
$Z = 47$

Kerala CEE - 2016

ATOMS

145264 The ratio of the surface area of the nuclei $_{52} {Te}^{125}$ to that of $_{13} {Al}^{27}$ is

1

\frac{5}{3}

2

\frac{125}{17}

3

\frac{1}{4}

4

\frac{25}{9}

5

\frac{3}{5}

Explanation:

D Given,
Atomic mass of $_{52} {Te}^{125} (A) = 125$
Atomic mass of $_{13} A l^{27} (A) = 27$
The radius of the nucleus
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
$R \propto (A)^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = {(\frac{125}{27})}^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$∵$ Surface area $A \propto R^{2}$
$∴ \frac{R_{Te}}{R_{Al}} = {(\frac{R_{Te}}{R_{Al}})}^{2} = \frac{R_{Te}}{R_{Al}} = {(\frac{5}{3})}^{2} = \frac{25}{9}$

Kerala CEE- 2014

ATOMS

145265 The ratio between the radii of nuclei with mass numbers 27 and 125 is

1

5 : 3

2

3 : 5

3

27 : 125

4

125 : 27

5

1 : 1

Explanation:

B Given,
$Mass (A_{1}) = 27$
$Mass (A_{2}) = 125$
We know that, radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant
$R =$ radius of nucleus
$R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{27}{125})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = \frac{3}{5}$
Hence, $R_{1} : R_{2} = 3 : 5$

Kerala CEE 2012

ATOMS

145266 The nuclear radius of a certain nucleus is $7.2$
$fm$ and it has charge of $1.28 \times 10^{- 17} C$ . The number of neutrons inside the nucleus is

1 136

2 142

3 140

4 132

5 126

Explanation:

A Given,
Radius of nucleus $(R) = 7.2 fm$
Charge of the nucleus $= 1.28 \times 10^{- 17} C$
We know that,
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} = 1.2 fm$
$A = {(\frac{R}{R_{0}})}^{3}$
$A = {(\frac{7.2}{1.2})}^{3} = (6)^{3} = 216$
Number of proton in the nucleus,
$Z = \frac{Charge of the nucleus}{Charge of a proton}$
$Z = \frac{1.28 \times 10^{- 17}}{1.6 \times 10^{- 19}}$
$Z = 80$
The number of neutrons in the nucleus $= A - Z$
$= 216 - 80$
$= 136$

Kerala CEE - 2008

ATOMS

145267 Radius of $_{2} {He}^{4}$ nucleus is 3 fermi. The radius of $_{82} {Pb}^{206}$ nucleus will be

1 5 fermi

2 6 fermi

3 11.16 fermi

4 8 fermi

Explanation:

C Given,
Radius of $_{2} {He}^{4}$ nucleus $(R_{1}) = 3 fm$
Radius of $_{82} {Pb}^{206}$ nucleus $(R_{2}) =$ ?
Mass of $_{2} {He}^{4} (A_{1}) = 4$
Mass of $_{82} {Pb}^{206} (A_{2}) = 206$
We know that,
$R = R_{0} (A)^{1 / 3}$
Here, $R \propto A^{1 / 3}$
$\frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3}$
$\frac{R_{2}}{3} = {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 3 \times {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 11.16 fm$

JCECE-2010

ATOMS

145263 If the ratio of radius of a nucleus with 61 neutrons to that of helium nucleus is 3 , then the atomic number of this nucleus is

1 27

2 47

3 51

4 61

5 108

Explanation:

B Given that,
Ratio of the radii $(\frac{R_{1}}{R_{2}}) = 3$
Number of neutrons $= (A - Z) = 61$
Where, $A =$ Atomic mass
$Z =$ Atomic number
We know that, radius of the nucleus $R = R_{0} (A)^{1 / 3}$
i.e., $R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$(3)^{3} = \frac{A_{1}}{A_{2}} \Rightarrow 27 = \frac{A_{1}}{A_{2}}$
$A_{1} = 27 \times A_{2}$
$A_{1} = 27 \times 4 {A_{2} =$ mass of helium $}$
$A_{1} = 108$
$∵ A - Z = 61$
$Z = A - 61 = 108 - 61 {∵ A_{1} = A}$
$Z = 47$

Kerala CEE - 2016

ATOMS

145264 The ratio of the surface area of the nuclei $_{52} {Te}^{125}$ to that of $_{13} {Al}^{27}$ is

1

\frac{5}{3}

2

\frac{125}{17}

3

\frac{1}{4}

4

\frac{25}{9}

5

\frac{3}{5}

Explanation:

D Given,
Atomic mass of $_{52} {Te}^{125} (A) = 125$
Atomic mass of $_{13} A l^{27} (A) = 27$
The radius of the nucleus
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
$R \propto (A)^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = {(\frac{125}{27})}^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$∵$ Surface area $A \propto R^{2}$
$∴ \frac{R_{Te}}{R_{Al}} = {(\frac{R_{Te}}{R_{Al}})}^{2} = \frac{R_{Te}}{R_{Al}} = {(\frac{5}{3})}^{2} = \frac{25}{9}$

Kerala CEE- 2014

ATOMS

145265 The ratio between the radii of nuclei with mass numbers 27 and 125 is

1

5 : 3

2

3 : 5

3

27 : 125

4

125 : 27

5

1 : 1

Explanation:

B Given,
$Mass (A_{1}) = 27$
$Mass (A_{2}) = 125$
We know that, radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant
$R =$ radius of nucleus
$R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{27}{125})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = \frac{3}{5}$
Hence, $R_{1} : R_{2} = 3 : 5$

Kerala CEE 2012

ATOMS

145266 The nuclear radius of a certain nucleus is $7.2$
$fm$ and it has charge of $1.28 \times 10^{- 17} C$ . The number of neutrons inside the nucleus is

1 136

2 142

3 140

4 132

5 126

Explanation:

A Given,
Radius of nucleus $(R) = 7.2 fm$
Charge of the nucleus $= 1.28 \times 10^{- 17} C$
We know that,
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} = 1.2 fm$
$A = {(\frac{R}{R_{0}})}^{3}$
$A = {(\frac{7.2}{1.2})}^{3} = (6)^{3} = 216$
Number of proton in the nucleus,
$Z = \frac{Charge of the nucleus}{Charge of a proton}$
$Z = \frac{1.28 \times 10^{- 17}}{1.6 \times 10^{- 19}}$
$Z = 80$
The number of neutrons in the nucleus $= A - Z$
$= 216 - 80$
$= 136$

Kerala CEE - 2008

ATOMS

145267 Radius of $_{2} {He}^{4}$ nucleus is 3 fermi. The radius of $_{82} {Pb}^{206}$ nucleus will be

1 5 fermi

2 6 fermi

3 11.16 fermi

4 8 fermi

Explanation:

C Given,
Radius of $_{2} {He}^{4}$ nucleus $(R_{1}) = 3 fm$
Radius of $_{82} {Pb}^{206}$ nucleus $(R_{2}) =$ ?
Mass of $_{2} {He}^{4} (A_{1}) = 4$
Mass of $_{82} {Pb}^{206} (A_{2}) = 206$
We know that,
$R = R_{0} (A)^{1 / 3}$
Here, $R \propto A^{1 / 3}$
$\frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3}$
$\frac{R_{2}}{3} = {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 3 \times {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 11.16 fm$

JCECE-2010

ATOMS

145263 If the ratio of radius of a nucleus with 61 neutrons to that of helium nucleus is 3 , then the atomic number of this nucleus is

1 27

2 47

3 51

4 61

5 108

Explanation:

B Given that,
Ratio of the radii $(\frac{R_{1}}{R_{2}}) = 3$
Number of neutrons $= (A - Z) = 61$
Where, $A =$ Atomic mass
$Z =$ Atomic number
We know that, radius of the nucleus $R = R_{0} (A)^{1 / 3}$
i.e., $R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$(3)^{3} = \frac{A_{1}}{A_{2}} \Rightarrow 27 = \frac{A_{1}}{A_{2}}$
$A_{1} = 27 \times A_{2}$
$A_{1} = 27 \times 4 {A_{2} =$ mass of helium $}$
$A_{1} = 108$
$∵ A - Z = 61$
$Z = A - 61 = 108 - 61 {∵ A_{1} = A}$
$Z = 47$

Kerala CEE - 2016

ATOMS

145264 The ratio of the surface area of the nuclei $_{52} {Te}^{125}$ to that of $_{13} {Al}^{27}$ is

1

\frac{5}{3}

2

\frac{125}{17}

3

\frac{1}{4}

4

\frac{25}{9}

5

\frac{3}{5}

Explanation:

D Given,
Atomic mass of $_{52} {Te}^{125} (A) = 125$
Atomic mass of $_{13} A l^{27} (A) = 27$
The radius of the nucleus
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
$R \propto (A)^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = {(\frac{125}{27})}^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$∵$ Surface area $A \propto R^{2}$
$∴ \frac{R_{Te}}{R_{Al}} = {(\frac{R_{Te}}{R_{Al}})}^{2} = \frac{R_{Te}}{R_{Al}} = {(\frac{5}{3})}^{2} = \frac{25}{9}$

Kerala CEE- 2014

ATOMS

145265 The ratio between the radii of nuclei with mass numbers 27 and 125 is

1

5 : 3

2

3 : 5

3

27 : 125

4

125 : 27

5

1 : 1

Explanation:

B Given,
$Mass (A_{1}) = 27$
$Mass (A_{2}) = 125$
We know that, radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant
$R =$ radius of nucleus
$R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{27}{125})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = \frac{3}{5}$
Hence, $R_{1} : R_{2} = 3 : 5$

Kerala CEE 2012

ATOMS

145266 The nuclear radius of a certain nucleus is $7.2$
$fm$ and it has charge of $1.28 \times 10^{- 17} C$ . The number of neutrons inside the nucleus is

1 136

2 142

3 140

4 132

5 126

Explanation:

A Given,
Radius of nucleus $(R) = 7.2 fm$
Charge of the nucleus $= 1.28 \times 10^{- 17} C$
We know that,
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} = 1.2 fm$
$A = {(\frac{R}{R_{0}})}^{3}$
$A = {(\frac{7.2}{1.2})}^{3} = (6)^{3} = 216$
Number of proton in the nucleus,
$Z = \frac{Charge of the nucleus}{Charge of a proton}$
$Z = \frac{1.28 \times 10^{- 17}}{1.6 \times 10^{- 19}}$
$Z = 80$
The number of neutrons in the nucleus $= A - Z$
$= 216 - 80$
$= 136$

Kerala CEE - 2008

ATOMS

145267 Radius of $_{2} {He}^{4}$ nucleus is 3 fermi. The radius of $_{82} {Pb}^{206}$ nucleus will be

1 5 fermi

2 6 fermi

3 11.16 fermi

4 8 fermi

Explanation:

C Given,
Radius of $_{2} {He}^{4}$ nucleus $(R_{1}) = 3 fm$
Radius of $_{82} {Pb}^{206}$ nucleus $(R_{2}) =$ ?
Mass of $_{2} {He}^{4} (A_{1}) = 4$
Mass of $_{82} {Pb}^{206} (A_{2}) = 206$
We know that,
$R = R_{0} (A)^{1 / 3}$
Here, $R \propto A^{1 / 3}$
$\frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3}$
$\frac{R_{2}}{3} = {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 3 \times {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 11.16 fm$

JCECE-2010

ATOMS

145263 If the ratio of radius of a nucleus with 61 neutrons to that of helium nucleus is 3 , then the atomic number of this nucleus is

1 27

2 47

3 51

4 61

5 108

Explanation:

B Given that,
Ratio of the radii $(\frac{R_{1}}{R_{2}}) = 3$
Number of neutrons $= (A - Z) = 61$
Where, $A =$ Atomic mass
$Z =$ Atomic number
We know that, radius of the nucleus $R = R_{0} (A)^{1 / 3}$
i.e., $R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$(3)^{3} = \frac{A_{1}}{A_{2}} \Rightarrow 27 = \frac{A_{1}}{A_{2}}$
$A_{1} = 27 \times A_{2}$
$A_{1} = 27 \times 4 {A_{2} =$ mass of helium $}$
$A_{1} = 108$
$∵ A - Z = 61$
$Z = A - 61 = 108 - 61 {∵ A_{1} = A}$
$Z = 47$

Kerala CEE - 2016

ATOMS

145264 The ratio of the surface area of the nuclei $_{52} {Te}^{125}$ to that of $_{13} {Al}^{27}$ is

1

\frac{5}{3}

2

\frac{125}{17}

3

\frac{1}{4}

4

\frac{25}{9}

5

\frac{3}{5}

Explanation:

D Given,
Atomic mass of $_{52} {Te}^{125} (A) = 125$
Atomic mass of $_{13} A l^{27} (A) = 27$
The radius of the nucleus
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
$R \propto (A)^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = {(\frac{125}{27})}^{1 / 3}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$∵$ Surface area $A \propto R^{2}$
$∴ \frac{R_{Te}}{R_{Al}} = {(\frac{R_{Te}}{R_{Al}})}^{2} = \frac{R_{Te}}{R_{Al}} = {(\frac{5}{3})}^{2} = \frac{25}{9}$

Kerala CEE- 2014

ATOMS

145265 The ratio between the radii of nuclei with mass numbers 27 and 125 is

1

5 : 3

2

3 : 5

3

27 : 125

4

125 : 27

5

1 : 1

Explanation:

B Given,
$Mass (A_{1}) = 27$
$Mass (A_{2}) = 125$
We know that, radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant
$R =$ radius of nucleus
$R \propto (A)^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = {(\frac{27}{125})}^{1 / 3}$
$\frac{R_{1}}{R_{2}} = \frac{3}{5}$
Hence, $R_{1} : R_{2} = 3 : 5$

Kerala CEE 2012

ATOMS

145266 The nuclear radius of a certain nucleus is $7.2$
$fm$ and it has charge of $1.28 \times 10^{- 17} C$ . The number of neutrons inside the nucleus is

1 136

2 142

3 140

4 132

5 126

Explanation:

A Given,
Radius of nucleus $(R) = 7.2 fm$
Charge of the nucleus $= 1.28 \times 10^{- 17} C$
We know that,
$R = R_{0} (A)^{1 / 3}$
Where $R_{0} = 1.2 fm$
$A = {(\frac{R}{R_{0}})}^{3}$
$A = {(\frac{7.2}{1.2})}^{3} = (6)^{3} = 216$
Number of proton in the nucleus,
$Z = \frac{Charge of the nucleus}{Charge of a proton}$
$Z = \frac{1.28 \times 10^{- 17}}{1.6 \times 10^{- 19}}$
$Z = 80$
The number of neutrons in the nucleus $= A - Z$
$= 216 - 80$
$= 136$

Kerala CEE - 2008

ATOMS

145267 Radius of $_{2} {He}^{4}$ nucleus is 3 fermi. The radius of $_{82} {Pb}^{206}$ nucleus will be

1 5 fermi

2 6 fermi

3 11.16 fermi

4 8 fermi

Explanation:

C Given,
Radius of $_{2} {He}^{4}$ nucleus $(R_{1}) = 3 fm$
Radius of $_{82} {Pb}^{206}$ nucleus $(R_{2}) =$ ?
Mass of $_{2} {He}^{4} (A_{1}) = 4$
Mass of $_{82} {Pb}^{206} (A_{2}) = 206$
We know that,
$R = R_{0} (A)^{1 / 3}$
Here, $R \propto A^{1 / 3}$
$\frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3}$
$\frac{R_{2}}{3} = {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 3 \times {(\frac{206}{4})}^{1 / 3}$
$R_{2} = 11.16 fm$

JCECE-2010