QBManager

NUCLEAR PHYSICS

147427 Atomic weight of boron is 10.81 and it has two isotopes $_{5}^{10} B$ and $_{5}^{11} B$ . Then, the ratio of atoms of $_{5}^{10} B$ and $_{5}^{11} B$ in nature would be

1 19:81

2

10 : 11

3

15 : 16

4

81 : 19

Explanation:

A Given that,
Atomic weight of boron $= 10.81$
Let, the percentage of $_{5} B^{10}$ atoms be $x$ and percentage of $_{5} B^{11}$ is $(100 - x)$ then the average atomic weight,
$\frac{10 x + 11 (100 - x)}{100} = 10.81$
$10 x + 1100 - 11 x = 1081$
$x = 19$
So, percentage of $_{5} B^{11} = 100 - 19 = 81$
Thus, the required ratio is -
$\frac{_{5} B^{10}}{_{5} B^{11}} = \frac{19}{81}$

AIPMT- 1998

NUCLEAR PHYSICS

147428 In compound $X (n, α) \to_{3} {Li}^{7}$ , the element $X$ is

1

_{2} {He}^{4}

2

_{5} B^{10}

3

_{5} B^{9}

4

_{4} B^{11}

Explanation:

B $_{Z}^{A} X +_{0} n^{1} \to_{3} {Li}^{7} +_{2} {He}^{4}$
For mass,
$A + 1 = 7 + 4$
$A = 10$
For Atomic number
$Z + 0 = 3 + 2$
$Z = 5$
Then, the element become $_{5} X^{10} =_{5} B^{10}$

AIPMT- 2001

NUCLEAR PHYSICS

147430 The volume occupied by an atom is greater than the volume of the nucleus by factor of about

1

10^{10}

2

10^{15}

3

10^{1}

4

10^{5}

Explanation:

B We know that,
Radius of atom is of the order of $(r_{1}) = 10^{- 10} m$
Radius of the nucleus is of the order of $(r_{2}) = 10^{- 15} m$
Then the ratio of required volume
$= \frac{Volume of atom}{Volume of nucleus} = \frac{\frac{4}{3} π r_{1}^{3}}{\frac{4}{3} π r_{2}^{3}}$
$= {(\frac{10^{- 10}}{10^{- 15}})}^{3}$
$= 10^{15}$

AIPMT- 2003

NUCLEAR PHYSICS

147431 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{Al}$ , then the radius of $_{53}^{125} Te$ nucleus is nearly

1

{(\frac{53}{13})}^{1 / 3} R_{Al}

2

\frac{5}{3} R_{Al}

3

\frac{3}{5} R_{Al}

4

{(\frac{13}{53})}^{1 / 3} R_{Al}

Explanation:

B Given, atomic mass of $Al$ and $Te$ are $A_{Al} = 27, A_{Te} = 125$ ,
We know that,
$R \propto A^{1 / 3}$
Where, $A =$ mass number
$R = radius of nucleus$
Ratio of radius of $Te$ and $Al$ is
$\frac{R_{Te}}{R_{Al}} = {(\frac{A_{Te}}{A_{Al}})}^{\frac{1}{3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{(125)^{1 / 3}}{(27)^{1 / 3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$R_{Te} = \frac{5}{3} R_{Al}$

AIPMT- 1992

NUCLEAR PHYSICS

147427 Atomic weight of boron is 10.81 and it has two isotopes $_{5}^{10} B$ and $_{5}^{11} B$ . Then, the ratio of atoms of $_{5}^{10} B$ and $_{5}^{11} B$ in nature would be

1 19:81

2

10 : 11

3

15 : 16

4

81 : 19

Explanation:

A Given that,
Atomic weight of boron $= 10.81$
Let, the percentage of $_{5} B^{10}$ atoms be $x$ and percentage of $_{5} B^{11}$ is $(100 - x)$ then the average atomic weight,
$\frac{10 x + 11 (100 - x)}{100} = 10.81$
$10 x + 1100 - 11 x = 1081$
$x = 19$
So, percentage of $_{5} B^{11} = 100 - 19 = 81$
Thus, the required ratio is -
$\frac{_{5} B^{10}}{_{5} B^{11}} = \frac{19}{81}$

AIPMT- 1998

NUCLEAR PHYSICS

147428 In compound $X (n, α) \to_{3} {Li}^{7}$ , the element $X$ is

1

_{2} {He}^{4}

2

_{5} B^{10}

3

_{5} B^{9}

4

_{4} B^{11}

Explanation:

B $_{Z}^{A} X +_{0} n^{1} \to_{3} {Li}^{7} +_{2} {He}^{4}$
For mass,
$A + 1 = 7 + 4$
$A = 10$
For Atomic number
$Z + 0 = 3 + 2$
$Z = 5$
Then, the element become $_{5} X^{10} =_{5} B^{10}$

AIPMT- 2001

NUCLEAR PHYSICS

147430 The volume occupied by an atom is greater than the volume of the nucleus by factor of about

1

10^{10}

2

10^{15}

3

10^{1}

4

10^{5}

Explanation:

B We know that,
Radius of atom is of the order of $(r_{1}) = 10^{- 10} m$
Radius of the nucleus is of the order of $(r_{2}) = 10^{- 15} m$
Then the ratio of required volume
$= \frac{Volume of atom}{Volume of nucleus} = \frac{\frac{4}{3} π r_{1}^{3}}{\frac{4}{3} π r_{2}^{3}}$
$= {(\frac{10^{- 10}}{10^{- 15}})}^{3}$
$= 10^{15}$

AIPMT- 2003

NUCLEAR PHYSICS

147431 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{Al}$ , then the radius of $_{53}^{125} Te$ nucleus is nearly

1

{(\frac{53}{13})}^{1 / 3} R_{Al}

2

\frac{5}{3} R_{Al}

3

\frac{3}{5} R_{Al}

4

{(\frac{13}{53})}^{1 / 3} R_{Al}

Explanation:

B Given, atomic mass of $Al$ and $Te$ are $A_{Al} = 27, A_{Te} = 125$ ,
We know that,
$R \propto A^{1 / 3}$
Where, $A =$ mass number
$R = radius of nucleus$
Ratio of radius of $Te$ and $Al$ is
$\frac{R_{Te}}{R_{Al}} = {(\frac{A_{Te}}{A_{Al}})}^{\frac{1}{3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{(125)^{1 / 3}}{(27)^{1 / 3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$R_{Te} = \frac{5}{3} R_{Al}$

AIPMT- 1992

NUCLEAR PHYSICS

147427 Atomic weight of boron is 10.81 and it has two isotopes $_{5}^{10} B$ and $_{5}^{11} B$ . Then, the ratio of atoms of $_{5}^{10} B$ and $_{5}^{11} B$ in nature would be

1 19:81

2

10 : 11

3

15 : 16

4

81 : 19

Explanation:

A Given that,
Atomic weight of boron $= 10.81$
Let, the percentage of $_{5} B^{10}$ atoms be $x$ and percentage of $_{5} B^{11}$ is $(100 - x)$ then the average atomic weight,
$\frac{10 x + 11 (100 - x)}{100} = 10.81$
$10 x + 1100 - 11 x = 1081$
$x = 19$
So, percentage of $_{5} B^{11} = 100 - 19 = 81$
Thus, the required ratio is -
$\frac{_{5} B^{10}}{_{5} B^{11}} = \frac{19}{81}$

AIPMT- 1998

NUCLEAR PHYSICS

147428 In compound $X (n, α) \to_{3} {Li}^{7}$ , the element $X$ is

1

_{2} {He}^{4}

2

_{5} B^{10}

3

_{5} B^{9}

4

_{4} B^{11}

Explanation:

B $_{Z}^{A} X +_{0} n^{1} \to_{3} {Li}^{7} +_{2} {He}^{4}$
For mass,
$A + 1 = 7 + 4$
$A = 10$
For Atomic number
$Z + 0 = 3 + 2$
$Z = 5$
Then, the element become $_{5} X^{10} =_{5} B^{10}$

AIPMT- 2001

NUCLEAR PHYSICS

147430 The volume occupied by an atom is greater than the volume of the nucleus by factor of about

1

10^{10}

2

10^{15}

3

10^{1}

4

10^{5}

Explanation:

B We know that,
Radius of atom is of the order of $(r_{1}) = 10^{- 10} m$
Radius of the nucleus is of the order of $(r_{2}) = 10^{- 15} m$
Then the ratio of required volume
$= \frac{Volume of atom}{Volume of nucleus} = \frac{\frac{4}{3} π r_{1}^{3}}{\frac{4}{3} π r_{2}^{3}}$
$= {(\frac{10^{- 10}}{10^{- 15}})}^{3}$
$= 10^{15}$

AIPMT- 2003

NUCLEAR PHYSICS

147431 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{Al}$ , then the radius of $_{53}^{125} Te$ nucleus is nearly

1

{(\frac{53}{13})}^{1 / 3} R_{Al}

2

\frac{5}{3} R_{Al}

3

\frac{3}{5} R_{Al}

4

{(\frac{13}{53})}^{1 / 3} R_{Al}

Explanation:

B Given, atomic mass of $Al$ and $Te$ are $A_{Al} = 27, A_{Te} = 125$ ,
We know that,
$R \propto A^{1 / 3}$
Where, $A =$ mass number
$R = radius of nucleus$
Ratio of radius of $Te$ and $Al$ is
$\frac{R_{Te}}{R_{Al}} = {(\frac{A_{Te}}{A_{Al}})}^{\frac{1}{3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{(125)^{1 / 3}}{(27)^{1 / 3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$R_{Te} = \frac{5}{3} R_{Al}$

AIPMT- 1992

NUCLEAR PHYSICS

147427 Atomic weight of boron is 10.81 and it has two isotopes $_{5}^{10} B$ and $_{5}^{11} B$ . Then, the ratio of atoms of $_{5}^{10} B$ and $_{5}^{11} B$ in nature would be

1 19:81

2

10 : 11

3

15 : 16

4

81 : 19

Explanation:

A Given that,
Atomic weight of boron $= 10.81$
Let, the percentage of $_{5} B^{10}$ atoms be $x$ and percentage of $_{5} B^{11}$ is $(100 - x)$ then the average atomic weight,
$\frac{10 x + 11 (100 - x)}{100} = 10.81$
$10 x + 1100 - 11 x = 1081$
$x = 19$
So, percentage of $_{5} B^{11} = 100 - 19 = 81$
Thus, the required ratio is -
$\frac{_{5} B^{10}}{_{5} B^{11}} = \frac{19}{81}$

AIPMT- 1998

NUCLEAR PHYSICS

147428 In compound $X (n, α) \to_{3} {Li}^{7}$ , the element $X$ is

1

_{2} {He}^{4}

2

_{5} B^{10}

3

_{5} B^{9}

4

_{4} B^{11}

Explanation:

B $_{Z}^{A} X +_{0} n^{1} \to_{3} {Li}^{7} +_{2} {He}^{4}$
For mass,
$A + 1 = 7 + 4$
$A = 10$
For Atomic number
$Z + 0 = 3 + 2$
$Z = 5$
Then, the element become $_{5} X^{10} =_{5} B^{10}$

AIPMT- 2001

NUCLEAR PHYSICS

147430 The volume occupied by an atom is greater than the volume of the nucleus by factor of about

1

10^{10}

2

10^{15}

3

10^{1}

4

10^{5}

Explanation:

B We know that,
Radius of atom is of the order of $(r_{1}) = 10^{- 10} m$
Radius of the nucleus is of the order of $(r_{2}) = 10^{- 15} m$
Then the ratio of required volume
$= \frac{Volume of atom}{Volume of nucleus} = \frac{\frac{4}{3} π r_{1}^{3}}{\frac{4}{3} π r_{2}^{3}}$
$= {(\frac{10^{- 10}}{10^{- 15}})}^{3}$
$= 10^{15}$

AIPMT- 2003

NUCLEAR PHYSICS

147431 If radius of the $_{13}^{27} Al$ nucleus is taken to be $R_{Al}$ , then the radius of $_{53}^{125} Te$ nucleus is nearly

1

{(\frac{53}{13})}^{1 / 3} R_{Al}

2

\frac{5}{3} R_{Al}

3

\frac{3}{5} R_{Al}

4

{(\frac{13}{53})}^{1 / 3} R_{Al}

Explanation:

B Given, atomic mass of $Al$ and $Te$ are $A_{Al} = 27, A_{Te} = 125$ ,
We know that,
$R \propto A^{1 / 3}$
Where, $A =$ mass number
$R = radius of nucleus$
Ratio of radius of $Te$ and $Al$ is
$\frac{R_{Te}}{R_{Al}} = {(\frac{A_{Te}}{A_{Al}})}^{\frac{1}{3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{(125)^{1 / 3}}{(27)^{1 / 3}}$
$\frac{R_{Te}}{R_{Al}} = \frac{5}{3}$
$R_{Te} = \frac{5}{3} R_{Al}$

AIPMT- 1992

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