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CHXI02:STRUCTURE OF ATOM

307109 Atomic radius is in the order of $1 0^{- 8} c m$ and nuclear radius is in the order of $1 0^{- 13} c m$ . Calculate what fraction of atom is occupied by nucleus?

1

1 0^{- 12}

2

1 0^{- 13}

3

1 0^{- 14}

4

1 0^{- 15}

Explanation:

Volume of nucleus $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3}) π \times (1 0^{- 13})^{3} c m^{3}$
Volume of atom
$= (\frac{4}{3}) π r^{3} = (\frac{4}{3}) π \times (1 0^{- 8})^{3} c m^{3}$
$∴ \frac{V_{n u c l e u s}}{V_{a t o m}} = \frac{1 0^{- 39}}{1 0^{- 24}} = 1 0^{- 15}$
$V_{n u c l e u s} = 1 0^{- 15} \times V_{a t o m}$

CHXI02:STRUCTURE OF ATOM

307110 The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number $A$ is given by $1.25 \times 10^{- 13} \times A^{1 / 3} cm$ . Radius of atom is $1 .0 \overset{^{\circ}}{A}$ . If the mass number is 64 then the dfraction of the atomic volume that is occupied by the nucleus is

1

1.0 \times 10^{- 3}

2

5.0 \times 10^{- 5}

3

2.5 \times 10^{- 2}

4

1.25 \times 10^{- 13}

Explanation:

Radius of nucleus
$= 1.25 \times 10^{- 13} \times A^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 64^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 4 c m = 5 \times 10^{- 13} c m$
Radius of atom = $= 1$ $A^{\circ}$ $= 10^{- 8} c m$
Volume of nucleus / volume of atom
$= \frac{4 / 3 π r^{3}}{4 / 3 π {(r_{A})}^{3}}$
Where, r = radius of nucleus
$r_{A} = r a d i u s o f a t o m$
$= \frac{(4 / 3) π {(5 \times 10^{- 13})}^{3}}{(4 / 3) π {(10^{- 8})}^{3}} = \frac{125 \times 10^{- 39}}{10^{- 24}}$
$= 125 \times 10^{- 15}$
$= 1.25 \times 10^{- 13}$

CHXI02:STRUCTURE OF ATOM

307111 Statement A :
Most of the space in the atoms is empty.
Statement B :
Most of the $α -$ particles passed through the foil remain undeflected.

1 Statement A is correct but statement B is incorrect

2 Statement A is incorrect but statement B is correct

3 Both statements are correct

4 Both statements are incorrect

Explanation:

Most of the $α$ -particles passed through the foil remain undeflected as most of the space in the atoms is empty.

CHXI02:STRUCTURE OF ATOM

307112 The density of fluorine nucleus in $g c m^{- 3}$ if its radius is $5 \times 10^{- 13}$ is ____. $(M a s s o f F = 19 a m u) .$ [Assuming the shape of the nucleus is spherical]

1

6 \times 10^{- 12}

2

6 \times 10^{10}

3

6 \times 10^{11}

4

6 \times 10^{12}

Explanation:

Mass of the nucleus of F atom $= 19 \times 1.66 \times 10^{- 24} g$
$(1 a m u = 1.660 \times 10^{24} g)$
Volume of the nucleus of F atom $= \frac{4}{3} π r^{3}$
$= \frac{4}{3} \times 3.14 {(5 \times 10^{- 13})}^{3}$
$= 525 \times 10^{- 39} c m^{3}$
Density of the nucleus of F atom $= \frac{M a s s}{V o l u m e}$
$= \frac{19 \times 1.66 \times 10^{- 24} g}{525 \times 10^{- 39} c m^{3}}$
$= 6.0 \times 10^{13} g c m^{- 3}$
$= 6 \times 10^{12}$

CHXI02:STRUCTURE OF ATOM

307109 Atomic radius is in the order of $1 0^{- 8} c m$ and nuclear radius is in the order of $1 0^{- 13} c m$ . Calculate what fraction of atom is occupied by nucleus?

1

1 0^{- 12}

2

1 0^{- 13}

3

1 0^{- 14}

4

1 0^{- 15}

Explanation:

Volume of nucleus $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3}) π \times (1 0^{- 13})^{3} c m^{3}$
Volume of atom
$= (\frac{4}{3}) π r^{3} = (\frac{4}{3}) π \times (1 0^{- 8})^{3} c m^{3}$
$∴ \frac{V_{n u c l e u s}}{V_{a t o m}} = \frac{1 0^{- 39}}{1 0^{- 24}} = 1 0^{- 15}$
$V_{n u c l e u s} = 1 0^{- 15} \times V_{a t o m}$

CHXI02:STRUCTURE OF ATOM

307110 The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number $A$ is given by $1.25 \times 10^{- 13} \times A^{1 / 3} cm$ . Radius of atom is $1 .0 \overset{^{\circ}}{A}$ . If the mass number is 64 then the dfraction of the atomic volume that is occupied by the nucleus is

1

1.0 \times 10^{- 3}

2

5.0 \times 10^{- 5}

3

2.5 \times 10^{- 2}

4

1.25 \times 10^{- 13}

Explanation:

Radius of nucleus
$= 1.25 \times 10^{- 13} \times A^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 64^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 4 c m = 5 \times 10^{- 13} c m$
Radius of atom = $= 1$ $A^{\circ}$ $= 10^{- 8} c m$
Volume of nucleus / volume of atom
$= \frac{4 / 3 π r^{3}}{4 / 3 π {(r_{A})}^{3}}$
Where, r = radius of nucleus
$r_{A} = r a d i u s o f a t o m$
$= \frac{(4 / 3) π {(5 \times 10^{- 13})}^{3}}{(4 / 3) π {(10^{- 8})}^{3}} = \frac{125 \times 10^{- 39}}{10^{- 24}}$
$= 125 \times 10^{- 15}$
$= 1.25 \times 10^{- 13}$

CHXI02:STRUCTURE OF ATOM

307111 Statement A :
Most of the space in the atoms is empty.
Statement B :
Most of the $α -$ particles passed through the foil remain undeflected.

1 Statement A is correct but statement B is incorrect

2 Statement A is incorrect but statement B is correct

3 Both statements are correct

4 Both statements are incorrect

Explanation:

Most of the $α$ -particles passed through the foil remain undeflected as most of the space in the atoms is empty.

CHXI02:STRUCTURE OF ATOM

307112 The density of fluorine nucleus in $g c m^{- 3}$ if its radius is $5 \times 10^{- 13}$ is ____. $(M a s s o f F = 19 a m u) .$ [Assuming the shape of the nucleus is spherical]

1

6 \times 10^{- 12}

2

6 \times 10^{10}

3

6 \times 10^{11}

4

6 \times 10^{12}

Explanation:

Mass of the nucleus of F atom $= 19 \times 1.66 \times 10^{- 24} g$
$(1 a m u = 1.660 \times 10^{24} g)$
Volume of the nucleus of F atom $= \frac{4}{3} π r^{3}$
$= \frac{4}{3} \times 3.14 {(5 \times 10^{- 13})}^{3}$
$= 525 \times 10^{- 39} c m^{3}$
Density of the nucleus of F atom $= \frac{M a s s}{V o l u m e}$
$= \frac{19 \times 1.66 \times 10^{- 24} g}{525 \times 10^{- 39} c m^{3}}$
$= 6.0 \times 10^{13} g c m^{- 3}$
$= 6 \times 10^{12}$

CHXI02:STRUCTURE OF ATOM

307109 Atomic radius is in the order of $1 0^{- 8} c m$ and nuclear radius is in the order of $1 0^{- 13} c m$ . Calculate what fraction of atom is occupied by nucleus?

1

1 0^{- 12}

2

1 0^{- 13}

3

1 0^{- 14}

4

1 0^{- 15}

Explanation:

Volume of nucleus $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3}) π \times (1 0^{- 13})^{3} c m^{3}$
Volume of atom
$= (\frac{4}{3}) π r^{3} = (\frac{4}{3}) π \times (1 0^{- 8})^{3} c m^{3}$
$∴ \frac{V_{n u c l e u s}}{V_{a t o m}} = \frac{1 0^{- 39}}{1 0^{- 24}} = 1 0^{- 15}$
$V_{n u c l e u s} = 1 0^{- 15} \times V_{a t o m}$

CHXI02:STRUCTURE OF ATOM

307110 The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number $A$ is given by $1.25 \times 10^{- 13} \times A^{1 / 3} cm$ . Radius of atom is $1 .0 \overset{^{\circ}}{A}$ . If the mass number is 64 then the dfraction of the atomic volume that is occupied by the nucleus is

1

1.0 \times 10^{- 3}

2

5.0 \times 10^{- 5}

3

2.5 \times 10^{- 2}

4

1.25 \times 10^{- 13}

Explanation:

Radius of nucleus
$= 1.25 \times 10^{- 13} \times A^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 64^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 4 c m = 5 \times 10^{- 13} c m$
Radius of atom = $= 1$ $A^{\circ}$ $= 10^{- 8} c m$
Volume of nucleus / volume of atom
$= \frac{4 / 3 π r^{3}}{4 / 3 π {(r_{A})}^{3}}$
Where, r = radius of nucleus
$r_{A} = r a d i u s o f a t o m$
$= \frac{(4 / 3) π {(5 \times 10^{- 13})}^{3}}{(4 / 3) π {(10^{- 8})}^{3}} = \frac{125 \times 10^{- 39}}{10^{- 24}}$
$= 125 \times 10^{- 15}$
$= 1.25 \times 10^{- 13}$

CHXI02:STRUCTURE OF ATOM

307111 Statement A :
Most of the space in the atoms is empty.
Statement B :
Most of the $α -$ particles passed through the foil remain undeflected.

1 Statement A is correct but statement B is incorrect

2 Statement A is incorrect but statement B is correct

3 Both statements are correct

4 Both statements are incorrect

Explanation:

Most of the $α$ -particles passed through the foil remain undeflected as most of the space in the atoms is empty.

CHXI02:STRUCTURE OF ATOM

307112 The density of fluorine nucleus in $g c m^{- 3}$ if its radius is $5 \times 10^{- 13}$ is ____. $(M a s s o f F = 19 a m u) .$ [Assuming the shape of the nucleus is spherical]

1

6 \times 10^{- 12}

2

6 \times 10^{10}

3

6 \times 10^{11}

4

6 \times 10^{12}

Explanation:

Mass of the nucleus of F atom $= 19 \times 1.66 \times 10^{- 24} g$
$(1 a m u = 1.660 \times 10^{24} g)$
Volume of the nucleus of F atom $= \frac{4}{3} π r^{3}$
$= \frac{4}{3} \times 3.14 {(5 \times 10^{- 13})}^{3}$
$= 525 \times 10^{- 39} c m^{3}$
Density of the nucleus of F atom $= \frac{M a s s}{V o l u m e}$
$= \frac{19 \times 1.66 \times 10^{- 24} g}{525 \times 10^{- 39} c m^{3}}$
$= 6.0 \times 10^{13} g c m^{- 3}$
$= 6 \times 10^{12}$

CHXI02:STRUCTURE OF ATOM

307109 Atomic radius is in the order of $1 0^{- 8} c m$ and nuclear radius is in the order of $1 0^{- 13} c m$ . Calculate what fraction of atom is occupied by nucleus?

1

1 0^{- 12}

2

1 0^{- 13}

3

1 0^{- 14}

4

1 0^{- 15}

Explanation:

Volume of nucleus $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3}) π \times (1 0^{- 13})^{3} c m^{3}$
Volume of atom
$= (\frac{4}{3}) π r^{3} = (\frac{4}{3}) π \times (1 0^{- 8})^{3} c m^{3}$
$∴ \frac{V_{n u c l e u s}}{V_{a t o m}} = \frac{1 0^{- 39}}{1 0^{- 24}} = 1 0^{- 15}$
$V_{n u c l e u s} = 1 0^{- 15} \times V_{a t o m}$

CHXI02:STRUCTURE OF ATOM

307110 The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number $A$ is given by $1.25 \times 10^{- 13} \times A^{1 / 3} cm$ . Radius of atom is $1 .0 \overset{^{\circ}}{A}$ . If the mass number is 64 then the dfraction of the atomic volume that is occupied by the nucleus is

1

1.0 \times 10^{- 3}

2

5.0 \times 10^{- 5}

3

2.5 \times 10^{- 2}

4

1.25 \times 10^{- 13}

Explanation:

Radius of nucleus
$= 1.25 \times 10^{- 13} \times A^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 64^{1 / 3} c m$
$= 1.25 \times 10^{- 13} \times 4 c m = 5 \times 10^{- 13} c m$
Radius of atom = $= 1$ $A^{\circ}$ $= 10^{- 8} c m$
Volume of nucleus / volume of atom
$= \frac{4 / 3 π r^{3}}{4 / 3 π {(r_{A})}^{3}}$
Where, r = radius of nucleus
$r_{A} = r a d i u s o f a t o m$
$= \frac{(4 / 3) π {(5 \times 10^{- 13})}^{3}}{(4 / 3) π {(10^{- 8})}^{3}} = \frac{125 \times 10^{- 39}}{10^{- 24}}$
$= 125 \times 10^{- 15}$
$= 1.25 \times 10^{- 13}$

CHXI02:STRUCTURE OF ATOM

307111 Statement A :
Most of the space in the atoms is empty.
Statement B :
Most of the $α -$ particles passed through the foil remain undeflected.

1 Statement A is correct but statement B is incorrect

2 Statement A is incorrect but statement B is correct

3 Both statements are correct

4 Both statements are incorrect

Explanation:

Most of the $α$ -particles passed through the foil remain undeflected as most of the space in the atoms is empty.

CHXI02:STRUCTURE OF ATOM

307112 The density of fluorine nucleus in $g c m^{- 3}$ if its radius is $5 \times 10^{- 13}$ is ____. $(M a s s o f F = 19 a m u) .$ [Assuming the shape of the nucleus is spherical]

1

6 \times 10^{- 12}

2

6 \times 10^{10}

3

6 \times 10^{11}

4

6 \times 10^{12}

Explanation:

Mass of the nucleus of F atom $= 19 \times 1.66 \times 10^{- 24} g$
$(1 a m u = 1.660 \times 10^{24} g)$
Volume of the nucleus of F atom $= \frac{4}{3} π r^{3}$
$= \frac{4}{3} \times 3.14 {(5 \times 10^{- 13})}^{3}$
$= 525 \times 10^{- 39} c m^{3}$
Density of the nucleus of F atom $= \frac{M a s s}{V o l u m e}$
$= \frac{19 \times 1.66 \times 10^{- 24} g}{525 \times 10^{- 39} c m^{3}}$
$= 6.0 \times 10^{13} g c m^{- 3}$
$= 6 \times 10^{12}$

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