QBManager

PHXII13:NUCLEI

363628 $M, M_{n} & M_{p}$ denotes the masses of a nucleus of $_{z} X^{A},$ a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then

1

M = (A - Z) M_{n} + Z M_{p}

2

M = Z M_{n} + (A - Z) M_{p}

3

M > (A - Z) M_{n} + Z M_{p}

4

M < (A - Z) M_{n} + Z M_{p}

PHXII13:NUCLEI

363629 One requires energy $E_{n}$ to remove a nucleon from a nucleus and an energy $E_{e}$ to remove an electron from the orbit of an atom. Then

1

E_{n} < E_{e}

2

E_{n} > E_{e}

3

E_{n} ⩾ E_{e}

4

E_{n} = E_{0}

PHXII13:NUCLEI

363630 $_{92}^{238} A \to_{90}^{234} B +_{2}^{4} D + Q$
In the given nuclear reaction, the approximate amount of energy released will be
[Given,mass of
$_{92}^{238} A = 238.05079 \times 931.5 M e V / c^{2}$
mass of $_{90}^{234} B = 234.04363 \times 931.5 M e V / c^{2},$
mass of $_{2}^{4} D = 4.00260 \times 931.5 M e V / c^{2}$

1

5.9 M e V

2

3.82 M e V

3

2.12 M e V

4

4.25 M e V

Explanation:

The energy released during the given reaction will be
$Δ Q = Δ m c^{2}$ where $Δ m =$ mass defect
$= (m_{A} - m_{B} - m_{D}) \times 931.5 M e V$
$= (238.05079 - 234.04363 - 4.00260) \times 931.5$
$= 4.25 M e V$

JEE - 2023

PHXII13:NUCLEI

363631 The mass defect in a particular nuclear reaction is $0.3 g$ . The amount of energy liberated (in $k W h$ ) is

1

1.56 \times 10^{6}

2

2.5 \times 10^{6}

3

3 \times 10^{6}

4

7.5 \times 10^{6}

Explanation:

According to mass energy equivalence,
$E = Δ m c^{2} = \frac{0.3}{1000} \times (3 \times 10^{8})^{2} = 2.7 \times 10^{13} J$
$= \frac{2.7 \times 10^{13}}{3.6 \times 10^{6}} = 7.5 \times 10^{6} k W h$

PHXII13:NUCLEI

363628 $M, M_{n} & M_{p}$ denotes the masses of a nucleus of $_{z} X^{A},$ a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then

1

M = (A - Z) M_{n} + Z M_{p}

2

M = Z M_{n} + (A - Z) M_{p}

3

M > (A - Z) M_{n} + Z M_{p}

4

M < (A - Z) M_{n} + Z M_{p}

PHXII13:NUCLEI

363629 One requires energy $E_{n}$ to remove a nucleon from a nucleus and an energy $E_{e}$ to remove an electron from the orbit of an atom. Then

1

E_{n} < E_{e}

2

E_{n} > E_{e}

3

E_{n} ⩾ E_{e}

4

E_{n} = E_{0}

PHXII13:NUCLEI

363630 $_{92}^{238} A \to_{90}^{234} B +_{2}^{4} D + Q$
In the given nuclear reaction, the approximate amount of energy released will be
[Given,mass of
$_{92}^{238} A = 238.05079 \times 931.5 M e V / c^{2}$
mass of $_{90}^{234} B = 234.04363 \times 931.5 M e V / c^{2},$
mass of $_{2}^{4} D = 4.00260 \times 931.5 M e V / c^{2}$

1

5.9 M e V

2

3.82 M e V

3

2.12 M e V

4

4.25 M e V

Explanation:

The energy released during the given reaction will be
$Δ Q = Δ m c^{2}$ where $Δ m =$ mass defect
$= (m_{A} - m_{B} - m_{D}) \times 931.5 M e V$
$= (238.05079 - 234.04363 - 4.00260) \times 931.5$
$= 4.25 M e V$

JEE - 2023

PHXII13:NUCLEI

363631 The mass defect in a particular nuclear reaction is $0.3 g$ . The amount of energy liberated (in $k W h$ ) is

1

1.56 \times 10^{6}

2

2.5 \times 10^{6}

3

3 \times 10^{6}

4

7.5 \times 10^{6}

Explanation:

According to mass energy equivalence,
$E = Δ m c^{2} = \frac{0.3}{1000} \times (3 \times 10^{8})^{2} = 2.7 \times 10^{13} J$
$= \frac{2.7 \times 10^{13}}{3.6 \times 10^{6}} = 7.5 \times 10^{6} k W h$

PHXII13:NUCLEI

363628 $M, M_{n} & M_{p}$ denotes the masses of a nucleus of $_{z} X^{A},$ a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then

1

M = (A - Z) M_{n} + Z M_{p}

2

M = Z M_{n} + (A - Z) M_{p}

3

M > (A - Z) M_{n} + Z M_{p}

4

M < (A - Z) M_{n} + Z M_{p}

PHXII13:NUCLEI

363629 One requires energy $E_{n}$ to remove a nucleon from a nucleus and an energy $E_{e}$ to remove an electron from the orbit of an atom. Then

1

E_{n} < E_{e}

2

E_{n} > E_{e}

3

E_{n} ⩾ E_{e}

4

E_{n} = E_{0}

PHXII13:NUCLEI

363630 $_{92}^{238} A \to_{90}^{234} B +_{2}^{4} D + Q$
In the given nuclear reaction, the approximate amount of energy released will be
[Given,mass of
$_{92}^{238} A = 238.05079 \times 931.5 M e V / c^{2}$
mass of $_{90}^{234} B = 234.04363 \times 931.5 M e V / c^{2},$
mass of $_{2}^{4} D = 4.00260 \times 931.5 M e V / c^{2}$

1

5.9 M e V

2

3.82 M e V

3

2.12 M e V

4

4.25 M e V

Explanation:

The energy released during the given reaction will be
$Δ Q = Δ m c^{2}$ where $Δ m =$ mass defect
$= (m_{A} - m_{B} - m_{D}) \times 931.5 M e V$
$= (238.05079 - 234.04363 - 4.00260) \times 931.5$
$= 4.25 M e V$

JEE - 2023

PHXII13:NUCLEI

363631 The mass defect in a particular nuclear reaction is $0.3 g$ . The amount of energy liberated (in $k W h$ ) is

1

1.56 \times 10^{6}

2

2.5 \times 10^{6}

3

3 \times 10^{6}

4

7.5 \times 10^{6}

Explanation:

According to mass energy equivalence,
$E = Δ m c^{2} = \frac{0.3}{1000} \times (3 \times 10^{8})^{2} = 2.7 \times 10^{13} J$
$= \frac{2.7 \times 10^{13}}{3.6 \times 10^{6}} = 7.5 \times 10^{6} k W h$

PHXII13:NUCLEI

363628 $M, M_{n} & M_{p}$ denotes the masses of a nucleus of $_{z} X^{A},$ a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then

1

M = (A - Z) M_{n} + Z M_{p}

2

M = Z M_{n} + (A - Z) M_{p}

3

M > (A - Z) M_{n} + Z M_{p}

4

M < (A - Z) M_{n} + Z M_{p}

PHXII13:NUCLEI

363629 One requires energy $E_{n}$ to remove a nucleon from a nucleus and an energy $E_{e}$ to remove an electron from the orbit of an atom. Then

1

E_{n} < E_{e}

2

E_{n} > E_{e}

3

E_{n} ⩾ E_{e}

4

E_{n} = E_{0}

PHXII13:NUCLEI

363630 $_{92}^{238} A \to_{90}^{234} B +_{2}^{4} D + Q$
In the given nuclear reaction, the approximate amount of energy released will be
[Given,mass of
$_{92}^{238} A = 238.05079 \times 931.5 M e V / c^{2}$
mass of $_{90}^{234} B = 234.04363 \times 931.5 M e V / c^{2},$
mass of $_{2}^{4} D = 4.00260 \times 931.5 M e V / c^{2}$

1

5.9 M e V

2

3.82 M e V

3

2.12 M e V

4

4.25 M e V

Explanation:

The energy released during the given reaction will be
$Δ Q = Δ m c^{2}$ where $Δ m =$ mass defect
$= (m_{A} - m_{B} - m_{D}) \times 931.5 M e V$
$= (238.05079 - 234.04363 - 4.00260) \times 931.5$
$= 4.25 M e V$

JEE - 2023

PHXII13:NUCLEI

363631 The mass defect in a particular nuclear reaction is $0.3 g$ . The amount of energy liberated (in $k W h$ ) is

1

1.56 \times 10^{6}

2

2.5 \times 10^{6}

3

3 \times 10^{6}

4

7.5 \times 10^{6}

Explanation:

According to mass energy equivalence,
$E = Δ m c^{2} = \frac{0.3}{1000} \times (3 \times 10^{8})^{2} = 2.7 \times 10^{13} J$
$= \frac{2.7 \times 10^{13}}{3.6 \times 10^{6}} = 7.5 \times 10^{6} k W h$

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