QBManager

PHXII13:NUCLEI

363619 The energy equivalent to a substance of mass 1 $g$ is

1

18 \times 10^{13} J

2

9 \times 10^{13} J

3

18 \times 10^{6} J

4

9 \times 10^{6} J

Explanation:

Here,
Mass, $m = 1, g = 10^{- 3} k g$
According to Einstein’s mass energy relation, the energy equivalent of mass $m$ is
$E = m c^{2}$ ( where $c$ is the speed of light in vaccum)
$= (10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 10^{- 3} \times 9 \times 10^{16} J = 9 \times 10^{13} J$

PHXII13:NUCLEI

363620 Let $m_{p}$ be the mass of a proton, $m_{n}$ the mass of a neturon, $M_{1}$ be the mass of $_{10} N e^{20}$ nucleus Then

1

M_{1} = 10 (m_{p} + m_{n})

2

M_{1} > 10 (m_{p} + m_{n})

3

M_{1} = 20 m_{n}

4

M_{1} < 10 (m_{p} + m_{n})

Explanation:

Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus $∴ M_{1} < 10 m_{p} + 10 m_{n}$

PHXII13:NUCLEI

363621 The atomic mass of $_{6} C^{12}$ is $12.000000 u$ and that of $_{6} C^{13}$ is $13.003354 u$ . The required energy to remove a neutron from $_{6} C^{13}$ , if mass of neutron is $1.008665 u$ , will be :

1

4.95 M e V

2

6.25 M e V

3

62.5 M e V

4

49.5 M e V

Explanation:

$_{6}^{13} C \to_{6}^{12} C +_{0}^{1} n$
$Δ m = 12 + 1.008665 - 13.003354$
$Δ m = 0.005311 a m u$
Using mass energy relation i.e.,
$u = Δ m \times 931.5 M e V$ ( $Δ m =$ mass defect $)$
$u = 4.947$
$\approx 4.95 M e V$

JEE - 2024

PHXII13:NUCLEI

363622 Energy equivalent to $21 g$ uranium is equal to:

1

9.0 \times 10^{13} J

2

21 \times 10^{- 9} J

3

189 \times 10^{13} J

4

1.9 \times 10^{12} J

Explanation:

$B . E = E = m c^{2} = (21 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$= 21 \times 10^{- 3} \times 9 \times 10^{16}$
$E = 189 \times 10^{13} J$ .

PHXII13:NUCLEI

363623 One microgram of matter converted into energy will be given

1

9 \times 10^{3} J

2

90 J

3

9 \times 10^{5} J

4

9 \times 10^{7} J

Explanation:

$E = Δ m c^{2} = 10^{- 9} \times (3 \times 10^{8})^{2} = 9 \times 10^{7} J$

PHXII13:NUCLEI

363619 The energy equivalent to a substance of mass 1 $g$ is

1

18 \times 10^{13} J

2

9 \times 10^{13} J

3

18 \times 10^{6} J

4

9 \times 10^{6} J

Explanation:

Here,
Mass, $m = 1, g = 10^{- 3} k g$
According to Einstein’s mass energy relation, the energy equivalent of mass $m$ is
$E = m c^{2}$ ( where $c$ is the speed of light in vaccum)
$= (10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 10^{- 3} \times 9 \times 10^{16} J = 9 \times 10^{13} J$

PHXII13:NUCLEI

363620 Let $m_{p}$ be the mass of a proton, $m_{n}$ the mass of a neturon, $M_{1}$ be the mass of $_{10} N e^{20}$ nucleus Then

1

M_{1} = 10 (m_{p} + m_{n})

2

M_{1} > 10 (m_{p} + m_{n})

3

M_{1} = 20 m_{n}

4

M_{1} < 10 (m_{p} + m_{n})

Explanation:

Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus $∴ M_{1} < 10 m_{p} + 10 m_{n}$

PHXII13:NUCLEI

363621 The atomic mass of $_{6} C^{12}$ is $12.000000 u$ and that of $_{6} C^{13}$ is $13.003354 u$ . The required energy to remove a neutron from $_{6} C^{13}$ , if mass of neutron is $1.008665 u$ , will be :

1

4.95 M e V

2

6.25 M e V

3

62.5 M e V

4

49.5 M e V

Explanation:

$_{6}^{13} C \to_{6}^{12} C +_{0}^{1} n$
$Δ m = 12 + 1.008665 - 13.003354$
$Δ m = 0.005311 a m u$
Using mass energy relation i.e.,
$u = Δ m \times 931.5 M e V$ ( $Δ m =$ mass defect $)$
$u = 4.947$
$\approx 4.95 M e V$

JEE - 2024

PHXII13:NUCLEI

363622 Energy equivalent to $21 g$ uranium is equal to:

1

9.0 \times 10^{13} J

2

21 \times 10^{- 9} J

3

189 \times 10^{13} J

4

1.9 \times 10^{12} J

Explanation:

$B . E = E = m c^{2} = (21 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$= 21 \times 10^{- 3} \times 9 \times 10^{16}$
$E = 189 \times 10^{13} J$ .

PHXII13:NUCLEI

363623 One microgram of matter converted into energy will be given

1

9 \times 10^{3} J

2

90 J

3

9 \times 10^{5} J

4

9 \times 10^{7} J

Explanation:

$E = Δ m c^{2} = 10^{- 9} \times (3 \times 10^{8})^{2} = 9 \times 10^{7} J$

PHXII13:NUCLEI

363619 The energy equivalent to a substance of mass 1 $g$ is

1

18 \times 10^{13} J

2

9 \times 10^{13} J

3

18 \times 10^{6} J

4

9 \times 10^{6} J

Explanation:

Here,
Mass, $m = 1, g = 10^{- 3} k g$
According to Einstein’s mass energy relation, the energy equivalent of mass $m$ is
$E = m c^{2}$ ( where $c$ is the speed of light in vaccum)
$= (10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 10^{- 3} \times 9 \times 10^{16} J = 9 \times 10^{13} J$

PHXII13:NUCLEI

363620 Let $m_{p}$ be the mass of a proton, $m_{n}$ the mass of a neturon, $M_{1}$ be the mass of $_{10} N e^{20}$ nucleus Then

1

M_{1} = 10 (m_{p} + m_{n})

2

M_{1} > 10 (m_{p} + m_{n})

3

M_{1} = 20 m_{n}

4

M_{1} < 10 (m_{p} + m_{n})

Explanation:

Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus $∴ M_{1} < 10 m_{p} + 10 m_{n}$

PHXII13:NUCLEI

363621 The atomic mass of $_{6} C^{12}$ is $12.000000 u$ and that of $_{6} C^{13}$ is $13.003354 u$ . The required energy to remove a neutron from $_{6} C^{13}$ , if mass of neutron is $1.008665 u$ , will be :

1

4.95 M e V

2

6.25 M e V

3

62.5 M e V

4

49.5 M e V

Explanation:

$_{6}^{13} C \to_{6}^{12} C +_{0}^{1} n$
$Δ m = 12 + 1.008665 - 13.003354$
$Δ m = 0.005311 a m u$
Using mass energy relation i.e.,
$u = Δ m \times 931.5 M e V$ ( $Δ m =$ mass defect $)$
$u = 4.947$
$\approx 4.95 M e V$

JEE - 2024

PHXII13:NUCLEI

363622 Energy equivalent to $21 g$ uranium is equal to:

1

9.0 \times 10^{13} J

2

21 \times 10^{- 9} J

3

189 \times 10^{13} J

4

1.9 \times 10^{12} J

Explanation:

$B . E = E = m c^{2} = (21 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$= 21 \times 10^{- 3} \times 9 \times 10^{16}$
$E = 189 \times 10^{13} J$ .

PHXII13:NUCLEI

363623 One microgram of matter converted into energy will be given

1

9 \times 10^{3} J

2

90 J

3

9 \times 10^{5} J

4

9 \times 10^{7} J

Explanation:

$E = Δ m c^{2} = 10^{- 9} \times (3 \times 10^{8})^{2} = 9 \times 10^{7} J$

PHXII13:NUCLEI

363619 The energy equivalent to a substance of mass 1 $g$ is

1

18 \times 10^{13} J

2

9 \times 10^{13} J

3

18 \times 10^{6} J

4

9 \times 10^{6} J

Explanation:

Here,
Mass, $m = 1, g = 10^{- 3} k g$
According to Einstein’s mass energy relation, the energy equivalent of mass $m$ is
$E = m c^{2}$ ( where $c$ is the speed of light in vaccum)
$= (10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 10^{- 3} \times 9 \times 10^{16} J = 9 \times 10^{13} J$

PHXII13:NUCLEI

363620 Let $m_{p}$ be the mass of a proton, $m_{n}$ the mass of a neturon, $M_{1}$ be the mass of $_{10} N e^{20}$ nucleus Then

1

M_{1} = 10 (m_{p} + m_{n})

2

M_{1} > 10 (m_{p} + m_{n})

3

M_{1} = 20 m_{n}

4

M_{1} < 10 (m_{p} + m_{n})

Explanation:

Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus $∴ M_{1} < 10 m_{p} + 10 m_{n}$

PHXII13:NUCLEI

363621 The atomic mass of $_{6} C^{12}$ is $12.000000 u$ and that of $_{6} C^{13}$ is $13.003354 u$ . The required energy to remove a neutron from $_{6} C^{13}$ , if mass of neutron is $1.008665 u$ , will be :

1

4.95 M e V

2

6.25 M e V

3

62.5 M e V

4

49.5 M e V

Explanation:

$_{6}^{13} C \to_{6}^{12} C +_{0}^{1} n$
$Δ m = 12 + 1.008665 - 13.003354$
$Δ m = 0.005311 a m u$
Using mass energy relation i.e.,
$u = Δ m \times 931.5 M e V$ ( $Δ m =$ mass defect $)$
$u = 4.947$
$\approx 4.95 M e V$

JEE - 2024

PHXII13:NUCLEI

363622 Energy equivalent to $21 g$ uranium is equal to:

1

9.0 \times 10^{13} J

2

21 \times 10^{- 9} J

3

189 \times 10^{13} J

4

1.9 \times 10^{12} J

Explanation:

$B . E = E = m c^{2} = (21 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$= 21 \times 10^{- 3} \times 9 \times 10^{16}$
$E = 189 \times 10^{13} J$ .

PHXII13:NUCLEI

363623 One microgram of matter converted into energy will be given

1

9 \times 10^{3} J

2

90 J

3

9 \times 10^{5} J

4

9 \times 10^{7} J

Explanation:

$E = Δ m c^{2} = 10^{- 9} \times (3 \times 10^{8})^{2} = 9 \times 10^{7} J$

PHXII13:NUCLEI

363619 The energy equivalent to a substance of mass 1 $g$ is

1

18 \times 10^{13} J

2

9 \times 10^{13} J

3

18 \times 10^{6} J

4

9 \times 10^{6} J

Explanation:

Here,
Mass, $m = 1, g = 10^{- 3} k g$
According to Einstein’s mass energy relation, the energy equivalent of mass $m$ is
$E = m c^{2}$ ( where $c$ is the speed of light in vaccum)
$= (10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 10^{- 3} \times 9 \times 10^{16} J = 9 \times 10^{13} J$

PHXII13:NUCLEI

363620 Let $m_{p}$ be the mass of a proton, $m_{n}$ the mass of a neturon, $M_{1}$ be the mass of $_{10} N e^{20}$ nucleus Then

1

M_{1} = 10 (m_{p} + m_{n})

2

M_{1} > 10 (m_{p} + m_{n})

3

M_{1} = 20 m_{n}

4

M_{1} < 10 (m_{p} + m_{n})

Explanation:

Since the mass of nucleus is less than that the sum of masses of nucleons forming the nucleus $∴ M_{1} < 10 m_{p} + 10 m_{n}$

PHXII13:NUCLEI

363621 The atomic mass of $_{6} C^{12}$ is $12.000000 u$ and that of $_{6} C^{13}$ is $13.003354 u$ . The required energy to remove a neutron from $_{6} C^{13}$ , if mass of neutron is $1.008665 u$ , will be :

1

4.95 M e V

2

6.25 M e V

3

62.5 M e V

4

49.5 M e V

Explanation:

$_{6}^{13} C \to_{6}^{12} C +_{0}^{1} n$
$Δ m = 12 + 1.008665 - 13.003354$
$Δ m = 0.005311 a m u$
Using mass energy relation i.e.,
$u = Δ m \times 931.5 M e V$ ( $Δ m =$ mass defect $)$
$u = 4.947$
$\approx 4.95 M e V$

JEE - 2024

PHXII13:NUCLEI

363622 Energy equivalent to $21 g$ uranium is equal to:

1

9.0 \times 10^{13} J

2

21 \times 10^{- 9} J

3

189 \times 10^{13} J

4

1.9 \times 10^{12} J

Explanation:

$B . E = E = m c^{2} = (21 \times 10^{- 3}) {(3 \times 10^{8})}^{2}$
$= 21 \times 10^{- 3} \times 9 \times 10^{16}$
$E = 189 \times 10^{13} J$ .

PHXII13:NUCLEI

363623 One microgram of matter converted into energy will be given

1

9 \times 10^{3} J

2

90 J

3

9 \times 10^{5} J

4

9 \times 10^{7} J

Explanation:

$E = Δ m c^{2} = 10^{- 9} \times (3 \times 10^{8})^{2} = 9 \times 10^{7} J$