QBManager

PHXII13:NUCLEI

363671 The binding energies of nuclei $X$ and $Y$ are $E_{1}$ and $E_{2}$ respecitvely. Two atoms of $X$ fuse to give one atom of $Y$ and an energy $Q$ is released. Then

1

Q = 2 E_{1} - E_{2}

2

Q = E_{2} - 2 E_{1}

3

Q < 2 E_{1} - E_{2}

4

Q > E_{2} - 2 E_{1}

Explanation:

During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released.
Hence , $Q = E_{2} - 2 E_{1}$

PHXII13:NUCLEI

363672 Binding energy of a certain nucleus is $18 \times 10^{8} J .$ How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?

1

0.2 μ g

2

10 μ g

3

2 μ g

4

20 μ g

Explanation:

The binding energy of nucleus, $B E = Δ m c^{2}$
$Δ m = \frac{B E}{c^{2}} = \frac{18 \times 10^{8}}{{(3 \times 10^{8})}^{2}} = \frac{18 \times 10^{8}}{9 \times 10^{16}}$
$= \frac{2}{10^{8}} k g = 2 \times 10^{- 8} k g$
$= 2 \times 10^{- 8} \times 1000 g$
$= 2 \times 10^{- 5} g = 20 μ g$

JEE - 2024

PHXII13:NUCLEI

363673 If $M_{0}$ is the mass of isotope $_{5}^{12} B, M_{p}$ and $M_{n}$ are the masses of proton and neutron, then nuclear binding energy of isotope is

1

(5 M_{p} + 7 M_{n} - M_{0}) c^{2}

2

(M_{0} - 5 M_{p}) c^{2}

3

(M_{0} - 12 M_{n}) c^{2}

4

(M_{0} - 5 M_{p} - 7 M_{n}) c^{2}

Explanation:

As we know,
$B . E . = Δ m c^{2}$
Here $Δ m = Z M_{p} + (A - Z) M_{n} - M_{0}$
$Δ m = (5 M_{p} + 7 M_{n} - M_{0})$
So, $B \cdot E . = (5 M_{p} + 7 M_{n} - M_{0}) c^{2}$

JEE - 2024

PHXII13:NUCLEI

363674 The mass defect of $_{2}^{4} H e$ is 0.03 $u$ . The binding energy per nucleon of helium(in $M e V$ ) is

1

69.825

2

6.9825

3

2.793

4

27.93

Explanation:

The mass defect of $_{2}^{4} H e, Δ m = 0.03 u$
The binding energy per nucleon $= \frac{Δ m \times 931}{4} M e V$
$= \frac{0.03 \times 931}{4} M e V = 6.9825 M e V$

KCET - 2017

PHXII13:NUCLEI

363671 The binding energies of nuclei $X$ and $Y$ are $E_{1}$ and $E_{2}$ respecitvely. Two atoms of $X$ fuse to give one atom of $Y$ and an energy $Q$ is released. Then

1

Q = 2 E_{1} - E_{2}

2

Q = E_{2} - 2 E_{1}

3

Q < 2 E_{1} - E_{2}

4

Q > E_{2} - 2 E_{1}

Explanation:

During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released.
Hence , $Q = E_{2} - 2 E_{1}$

PHXII13:NUCLEI

363672 Binding energy of a certain nucleus is $18 \times 10^{8} J .$ How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?

1

0.2 μ g

2

10 μ g

3

2 μ g

4

20 μ g

Explanation:

The binding energy of nucleus, $B E = Δ m c^{2}$
$Δ m = \frac{B E}{c^{2}} = \frac{18 \times 10^{8}}{{(3 \times 10^{8})}^{2}} = \frac{18 \times 10^{8}}{9 \times 10^{16}}$
$= \frac{2}{10^{8}} k g = 2 \times 10^{- 8} k g$
$= 2 \times 10^{- 8} \times 1000 g$
$= 2 \times 10^{- 5} g = 20 μ g$

JEE - 2024

PHXII13:NUCLEI

363673 If $M_{0}$ is the mass of isotope $_{5}^{12} B, M_{p}$ and $M_{n}$ are the masses of proton and neutron, then nuclear binding energy of isotope is

1

(5 M_{p} + 7 M_{n} - M_{0}) c^{2}

2

(M_{0} - 5 M_{p}) c^{2}

3

(M_{0} - 12 M_{n}) c^{2}

4

(M_{0} - 5 M_{p} - 7 M_{n}) c^{2}

Explanation:

As we know,
$B . E . = Δ m c^{2}$
Here $Δ m = Z M_{p} + (A - Z) M_{n} - M_{0}$
$Δ m = (5 M_{p} + 7 M_{n} - M_{0})$
So, $B \cdot E . = (5 M_{p} + 7 M_{n} - M_{0}) c^{2}$

JEE - 2024

PHXII13:NUCLEI

363674 The mass defect of $_{2}^{4} H e$ is 0.03 $u$ . The binding energy per nucleon of helium(in $M e V$ ) is

1

69.825

2

6.9825

3

2.793

4

27.93

Explanation:

The mass defect of $_{2}^{4} H e, Δ m = 0.03 u$
The binding energy per nucleon $= \frac{Δ m \times 931}{4} M e V$
$= \frac{0.03 \times 931}{4} M e V = 6.9825 M e V$

KCET - 2017

PHXII13:NUCLEI

363671 The binding energies of nuclei $X$ and $Y$ are $E_{1}$ and $E_{2}$ respecitvely. Two atoms of $X$ fuse to give one atom of $Y$ and an energy $Q$ is released. Then

1

Q = 2 E_{1} - E_{2}

2

Q = E_{2} - 2 E_{1}

3

Q < 2 E_{1} - E_{2}

4

Q > E_{2} - 2 E_{1}

Explanation:

During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released.
Hence , $Q = E_{2} - 2 E_{1}$

PHXII13:NUCLEI

363672 Binding energy of a certain nucleus is $18 \times 10^{8} J .$ How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?

1

0.2 μ g

2

10 μ g

3

2 μ g

4

20 μ g

Explanation:

The binding energy of nucleus, $B E = Δ m c^{2}$
$Δ m = \frac{B E}{c^{2}} = \frac{18 \times 10^{8}}{{(3 \times 10^{8})}^{2}} = \frac{18 \times 10^{8}}{9 \times 10^{16}}$
$= \frac{2}{10^{8}} k g = 2 \times 10^{- 8} k g$
$= 2 \times 10^{- 8} \times 1000 g$
$= 2 \times 10^{- 5} g = 20 μ g$

JEE - 2024

PHXII13:NUCLEI

363673 If $M_{0}$ is the mass of isotope $_{5}^{12} B, M_{p}$ and $M_{n}$ are the masses of proton and neutron, then nuclear binding energy of isotope is

1

(5 M_{p} + 7 M_{n} - M_{0}) c^{2}

2

(M_{0} - 5 M_{p}) c^{2}

3

(M_{0} - 12 M_{n}) c^{2}

4

(M_{0} - 5 M_{p} - 7 M_{n}) c^{2}

Explanation:

As we know,
$B . E . = Δ m c^{2}$
Here $Δ m = Z M_{p} + (A - Z) M_{n} - M_{0}$
$Δ m = (5 M_{p} + 7 M_{n} - M_{0})$
So, $B \cdot E . = (5 M_{p} + 7 M_{n} - M_{0}) c^{2}$

JEE - 2024

PHXII13:NUCLEI

363674 The mass defect of $_{2}^{4} H e$ is 0.03 $u$ . The binding energy per nucleon of helium(in $M e V$ ) is

1

69.825

2

6.9825

3

2.793

4

27.93

Explanation:

The mass defect of $_{2}^{4} H e, Δ m = 0.03 u$
The binding energy per nucleon $= \frac{Δ m \times 931}{4} M e V$
$= \frac{0.03 \times 931}{4} M e V = 6.9825 M e V$

KCET - 2017

PHXII13:NUCLEI

363671 The binding energies of nuclei $X$ and $Y$ are $E_{1}$ and $E_{2}$ respecitvely. Two atoms of $X$ fuse to give one atom of $Y$ and an energy $Q$ is released. Then

1

Q = 2 E_{1} - E_{2}

2

Q = E_{2} - 2 E_{1}

3

Q < 2 E_{1} - E_{2}

4

Q > E_{2} - 2 E_{1}

Explanation:

During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released.
Hence , $Q = E_{2} - 2 E_{1}$

PHXII13:NUCLEI

363672 Binding energy of a certain nucleus is $18 \times 10^{8} J .$ How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus?

1

0.2 μ g

2

10 μ g

3

2 μ g

4

20 μ g

Explanation:

The binding energy of nucleus, $B E = Δ m c^{2}$
$Δ m = \frac{B E}{c^{2}} = \frac{18 \times 10^{8}}{{(3 \times 10^{8})}^{2}} = \frac{18 \times 10^{8}}{9 \times 10^{16}}$
$= \frac{2}{10^{8}} k g = 2 \times 10^{- 8} k g$
$= 2 \times 10^{- 8} \times 1000 g$
$= 2 \times 10^{- 5} g = 20 μ g$

JEE - 2024

PHXII13:NUCLEI

363673 If $M_{0}$ is the mass of isotope $_{5}^{12} B, M_{p}$ and $M_{n}$ are the masses of proton and neutron, then nuclear binding energy of isotope is

1

(5 M_{p} + 7 M_{n} - M_{0}) c^{2}

2

(M_{0} - 5 M_{p}) c^{2}

3

(M_{0} - 12 M_{n}) c^{2}

4

(M_{0} - 5 M_{p} - 7 M_{n}) c^{2}

Explanation:

As we know,
$B . E . = Δ m c^{2}$
Here $Δ m = Z M_{p} + (A - Z) M_{n} - M_{0}$
$Δ m = (5 M_{p} + 7 M_{n} - M_{0})$
So, $B \cdot E . = (5 M_{p} + 7 M_{n} - M_{0}) c^{2}$

JEE - 2024

PHXII13:NUCLEI

363674 The mass defect of $_{2}^{4} H e$ is 0.03 $u$ . The binding energy per nucleon of helium(in $M e V$ ) is

1

69.825

2

6.9825

3

2.793

4

27.93

Explanation:

The mass defect of $_{2}^{4} H e, Δ m = 0.03 u$
The binding energy per nucleon $= \frac{Δ m \times 931}{4} M e V$
$= \frac{0.03 \times 931}{4} M e V = 6.9825 M e V$

KCET - 2017