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PHXII13:NUCLEI

363848 The radius of a copper nucleus is of the order of

1

10^{- 16} m

2

10^{- 14} m

3

10^{- 12} m

4

10^{- 9} m

Explanation:

Copper nucleus has a radius of the
order of $10^{- 14} m$

PHXII13:NUCLEI

363849 Radius of $_{2} H e^{4}$ nucleus is $3 F m$ . The radius of $_{82} P b^{206}$ nucleus will be:

1

5 F m

2

6 F m

3

11.16 F m

4

8 F m

Explanation:

We have, $r α A^{1 / 3}$
$\Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{206}{4})}^{1 / 3}$
$∴ r_{2} = 3 {(\frac{206}{4})}^{1 / 3} = 11.16 F m$

PHXII13:NUCLEI

363850 The nucleus which has radius one-third of the radius of $O s^{189}$ is:

1

B e^{9}

2

L i^{7}

3

F^{19}

4

C^{12}

Explanation:

The radius of a nucleus is given by the relation, $r = r_{0} (A)^{1 / 3}$
where, $r_{0} =$ constant
For $O s^{189}, A_{1} = 189, r_{1} = r$
Thus,using $r_{2} = \frac{r}{3}$ , we know that,
$\frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3} \Rightarrow 3 = {(\frac{189}{A_{2}})}^{1 / 3}$
$\Rightarrow 27 = \frac{189}{A_{2}} \Rightarrow A_{2} = 7$
Thus, the answer is $L i^{7}$

PHXII13:NUCLEI

363851 The ratio of the nuclear radii of elements with mass numbers 216 and 125 is

1

6 : 5

2

\sqrt{216} : \sqrt{125}

3

216 : 125

4 none of these

Explanation:

For a nucleus of mass number $A$ , its radius is given by
$R = R_{0} A^{1 / 3}$
Hence for two elements with $A = 216$ and $A = 125$ ,
$∴ \frac{R (^{216} X)}{R (^{125} Y)} = \frac{\sqrt[3]{216}}{\sqrt[3]{125}} = \frac{6}{5}$

KCET - 2008

PHXII13:NUCLEI

363848 The radius of a copper nucleus is of the order of

1

10^{- 16} m

2

10^{- 14} m

3

10^{- 12} m

4

10^{- 9} m

Explanation:

Copper nucleus has a radius of the
order of $10^{- 14} m$

PHXII13:NUCLEI

363849 Radius of $_{2} H e^{4}$ nucleus is $3 F m$ . The radius of $_{82} P b^{206}$ nucleus will be:

1

5 F m

2

6 F m

3

11.16 F m

4

8 F m

Explanation:

We have, $r α A^{1 / 3}$
$\Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{206}{4})}^{1 / 3}$
$∴ r_{2} = 3 {(\frac{206}{4})}^{1 / 3} = 11.16 F m$

PHXII13:NUCLEI

363850 The nucleus which has radius one-third of the radius of $O s^{189}$ is:

1

B e^{9}

2

L i^{7}

3

F^{19}

4

C^{12}

Explanation:

The radius of a nucleus is given by the relation, $r = r_{0} (A)^{1 / 3}$
where, $r_{0} =$ constant
For $O s^{189}, A_{1} = 189, r_{1} = r$
Thus,using $r_{2} = \frac{r}{3}$ , we know that,
$\frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3} \Rightarrow 3 = {(\frac{189}{A_{2}})}^{1 / 3}$
$\Rightarrow 27 = \frac{189}{A_{2}} \Rightarrow A_{2} = 7$
Thus, the answer is $L i^{7}$

PHXII13:NUCLEI

363851 The ratio of the nuclear radii of elements with mass numbers 216 and 125 is

1

6 : 5

2

\sqrt{216} : \sqrt{125}

3

216 : 125

4 none of these

Explanation:

For a nucleus of mass number $A$ , its radius is given by
$R = R_{0} A^{1 / 3}$
Hence for two elements with $A = 216$ and $A = 125$ ,
$∴ \frac{R (^{216} X)}{R (^{125} Y)} = \frac{\sqrt[3]{216}}{\sqrt[3]{125}} = \frac{6}{5}$

KCET - 2008

PHXII13:NUCLEI

363848 The radius of a copper nucleus is of the order of

1

10^{- 16} m

2

10^{- 14} m

3

10^{- 12} m

4

10^{- 9} m

Explanation:

Copper nucleus has a radius of the
order of $10^{- 14} m$

PHXII13:NUCLEI

363849 Radius of $_{2} H e^{4}$ nucleus is $3 F m$ . The radius of $_{82} P b^{206}$ nucleus will be:

1

5 F m

2

6 F m

3

11.16 F m

4

8 F m

Explanation:

We have, $r α A^{1 / 3}$
$\Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{206}{4})}^{1 / 3}$
$∴ r_{2} = 3 {(\frac{206}{4})}^{1 / 3} = 11.16 F m$

PHXII13:NUCLEI

363850 The nucleus which has radius one-third of the radius of $O s^{189}$ is:

1

B e^{9}

2

L i^{7}

3

F^{19}

4

C^{12}

Explanation:

The radius of a nucleus is given by the relation, $r = r_{0} (A)^{1 / 3}$
where, $r_{0} =$ constant
For $O s^{189}, A_{1} = 189, r_{1} = r$
Thus,using $r_{2} = \frac{r}{3}$ , we know that,
$\frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3} \Rightarrow 3 = {(\frac{189}{A_{2}})}^{1 / 3}$
$\Rightarrow 27 = \frac{189}{A_{2}} \Rightarrow A_{2} = 7$
Thus, the answer is $L i^{7}$

PHXII13:NUCLEI

363851 The ratio of the nuclear radii of elements with mass numbers 216 and 125 is

1

6 : 5

2

\sqrt{216} : \sqrt{125}

3

216 : 125

4 none of these

Explanation:

For a nucleus of mass number $A$ , its radius is given by
$R = R_{0} A^{1 / 3}$
Hence for two elements with $A = 216$ and $A = 125$ ,
$∴ \frac{R (^{216} X)}{R (^{125} Y)} = \frac{\sqrt[3]{216}}{\sqrt[3]{125}} = \frac{6}{5}$

KCET - 2008

PHXII13:NUCLEI

363848 The radius of a copper nucleus is of the order of

1

10^{- 16} m

2

10^{- 14} m

3

10^{- 12} m

4

10^{- 9} m

Explanation:

Copper nucleus has a radius of the
order of $10^{- 14} m$

PHXII13:NUCLEI

363849 Radius of $_{2} H e^{4}$ nucleus is $3 F m$ . The radius of $_{82} P b^{206}$ nucleus will be:

1

5 F m

2

6 F m

3

11.16 F m

4

8 F m

Explanation:

We have, $r α A^{1 / 3}$
$\Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{206}{4})}^{1 / 3}$
$∴ r_{2} = 3 {(\frac{206}{4})}^{1 / 3} = 11.16 F m$

PHXII13:NUCLEI

363850 The nucleus which has radius one-third of the radius of $O s^{189}$ is:

1

B e^{9}

2

L i^{7}

3

F^{19}

4

C^{12}

Explanation:

The radius of a nucleus is given by the relation, $r = r_{0} (A)^{1 / 3}$
where, $r_{0} =$ constant
For $O s^{189}, A_{1} = 189, r_{1} = r$
Thus,using $r_{2} = \frac{r}{3}$ , we know that,
$\frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3} \Rightarrow 3 = {(\frac{189}{A_{2}})}^{1 / 3}$
$\Rightarrow 27 = \frac{189}{A_{2}} \Rightarrow A_{2} = 7$
Thus, the answer is $L i^{7}$

PHXII13:NUCLEI

363851 The ratio of the nuclear radii of elements with mass numbers 216 and 125 is

1

6 : 5

2

\sqrt{216} : \sqrt{125}

3

216 : 125

4 none of these

Explanation:

For a nucleus of mass number $A$ , its radius is given by
$R = R_{0} A^{1 / 3}$
Hence for two elements with $A = 216$ and $A = 125$ ,
$∴ \frac{R (^{216} X)}{R (^{125} Y)} = \frac{\sqrt[3]{216}}{\sqrt[3]{125}} = \frac{6}{5}$

KCET - 2008

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