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PHXII13:NUCLEI

363632 The energy equivalent of $0.5 g$ of a substance is:

1

4.5 \times 10^{13} J

2

1.5 \times 10^{13} J

3

0.5 \times 10^{13} J

4

3.5 \times 10^{13} J

Explanation:

$E = m c^{2}$
$= 0.5 \times 10^{- 3} \times 9 \times 10^{16}$
$= 4.5 \times 10^{13} J$

NEET - 2020

PHXII13:NUCLEI

363633 The energy supplied by a power plant is 40 million kilowatt hour. It is supplied by annihilation of matter, the mass that is annihilated is

1

1.6 g

2

1.6 k g

3

1.6 m g

4

1.6 a m u

Explanation:

$E = M C^{2}, E = 40 \times 10^{6} \times 10^{3} \times 60 \times 60 J$ .
$M = \frac{40 \times 10^{6} \times 10^{3} \times 60 \times 60}{9 \times 10^{16}}$
$= \frac{4}{9} \times 36 \times 10^{- 4} = 16 \times 10^{- 4} k g = 1.6 g .$

PHXII13:NUCLEI

363634 Assume that a neutron breaks into a proton and an electron. The energy absorbed during this process is: (mass of neutron $= 1.6725 \times 10^{- 27} k g,$ mass of proton $= 1.6725 \times 10^{- 27} k g,$ mass of electron $= 9 \times 10^{- 31} k g$ ).

1

- 0.51 M e V

2

- 7.10 M e V

3

- 6.30 M e V

4

- 5.4 M e V

Explanation:

$_{0}^{1} n \to_{1}^{1} P +_{- 1} e^{0} + \bar{v} + Q$
The mass defect during the process
$Δ m = m_{n} - m_{H} - m_{e} = 1.6725 \times 10^{27}$
$- (1.6725 \times 10^{- 27} + 9 \times 10^{- 31} k g)$
$= - 9 \times 10^{- 31} k g$
Here $Δ m$ is $(-) v e$ so the reaction absorbs energy.The energy absorbed during the process
$E = Δ m c^{2}$
$E = 9 \times 10^{- 31} \times 9 \times 10^{16} = 81 \times 10^{- 15} joules$
$E = \frac{81 \times 10^{- 15}}{1.6 \times 10^{- 19}} = 0.511$ $M e v$
The correct answer is $- 0.511 M e V$

PHXII13:NUCLEI

363635 The difference between the mass of a nucleus and the combined mass of its nucleons is

1 Zero, positive or negative

2 Zero

3 Positive

4 Negative

PHXII13:NUCLEI

363636 $6.4 \times 10^{- 19}$ joule is approximately

1 4 electron volt

2 6 electron volt

3 8 electron volt

4 1 electron volt

Explanation:

$1 e V = 1.6 \times 10^{- 19} J$
$∴ 6.4 \times 10^{- 19} J = \frac{6.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 4 e V .$
So correct option is (1)

PHXII13:NUCLEI

363632 The energy equivalent of $0.5 g$ of a substance is:

1

4.5 \times 10^{13} J

2

1.5 \times 10^{13} J

3

0.5 \times 10^{13} J

4

3.5 \times 10^{13} J

Explanation:

$E = m c^{2}$
$= 0.5 \times 10^{- 3} \times 9 \times 10^{16}$
$= 4.5 \times 10^{13} J$

NEET - 2020

PHXII13:NUCLEI

363633 The energy supplied by a power plant is 40 million kilowatt hour. It is supplied by annihilation of matter, the mass that is annihilated is

1

1.6 g

2

1.6 k g

3

1.6 m g

4

1.6 a m u

Explanation:

$E = M C^{2}, E = 40 \times 10^{6} \times 10^{3} \times 60 \times 60 J$ .
$M = \frac{40 \times 10^{6} \times 10^{3} \times 60 \times 60}{9 \times 10^{16}}$
$= \frac{4}{9} \times 36 \times 10^{- 4} = 16 \times 10^{- 4} k g = 1.6 g .$

PHXII13:NUCLEI

363634 Assume that a neutron breaks into a proton and an electron. The energy absorbed during this process is: (mass of neutron $= 1.6725 \times 10^{- 27} k g,$ mass of proton $= 1.6725 \times 10^{- 27} k g,$ mass of electron $= 9 \times 10^{- 31} k g$ ).

1

- 0.51 M e V

2

- 7.10 M e V

3

- 6.30 M e V

4

- 5.4 M e V

Explanation:

$_{0}^{1} n \to_{1}^{1} P +_{- 1} e^{0} + \bar{v} + Q$
The mass defect during the process
$Δ m = m_{n} - m_{H} - m_{e} = 1.6725 \times 10^{27}$
$- (1.6725 \times 10^{- 27} + 9 \times 10^{- 31} k g)$
$= - 9 \times 10^{- 31} k g$
Here $Δ m$ is $(-) v e$ so the reaction absorbs energy.The energy absorbed during the process
$E = Δ m c^{2}$
$E = 9 \times 10^{- 31} \times 9 \times 10^{16} = 81 \times 10^{- 15} joules$
$E = \frac{81 \times 10^{- 15}}{1.6 \times 10^{- 19}} = 0.511$ $M e v$
The correct answer is $- 0.511 M e V$

PHXII13:NUCLEI

363635 The difference between the mass of a nucleus and the combined mass of its nucleons is

1 Zero, positive or negative

2 Zero

3 Positive

4 Negative

PHXII13:NUCLEI

363636 $6.4 \times 10^{- 19}$ joule is approximately

1 4 electron volt

2 6 electron volt

3 8 electron volt

4 1 electron volt

Explanation:

$1 e V = 1.6 \times 10^{- 19} J$
$∴ 6.4 \times 10^{- 19} J = \frac{6.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 4 e V .$
So correct option is (1)

PHXII13:NUCLEI

363632 The energy equivalent of $0.5 g$ of a substance is:

1

4.5 \times 10^{13} J

2

1.5 \times 10^{13} J

3

0.5 \times 10^{13} J

4

3.5 \times 10^{13} J

Explanation:

$E = m c^{2}$
$= 0.5 \times 10^{- 3} \times 9 \times 10^{16}$
$= 4.5 \times 10^{13} J$

NEET - 2020

PHXII13:NUCLEI

363633 The energy supplied by a power plant is 40 million kilowatt hour. It is supplied by annihilation of matter, the mass that is annihilated is

1

1.6 g

2

1.6 k g

3

1.6 m g

4

1.6 a m u

Explanation:

$E = M C^{2}, E = 40 \times 10^{6} \times 10^{3} \times 60 \times 60 J$ .
$M = \frac{40 \times 10^{6} \times 10^{3} \times 60 \times 60}{9 \times 10^{16}}$
$= \frac{4}{9} \times 36 \times 10^{- 4} = 16 \times 10^{- 4} k g = 1.6 g .$

PHXII13:NUCLEI

363634 Assume that a neutron breaks into a proton and an electron. The energy absorbed during this process is: (mass of neutron $= 1.6725 \times 10^{- 27} k g,$ mass of proton $= 1.6725 \times 10^{- 27} k g,$ mass of electron $= 9 \times 10^{- 31} k g$ ).

1

- 0.51 M e V

2

- 7.10 M e V

3

- 6.30 M e V

4

- 5.4 M e V

Explanation:

$_{0}^{1} n \to_{1}^{1} P +_{- 1} e^{0} + \bar{v} + Q$
The mass defect during the process
$Δ m = m_{n} - m_{H} - m_{e} = 1.6725 \times 10^{27}$
$- (1.6725 \times 10^{- 27} + 9 \times 10^{- 31} k g)$
$= - 9 \times 10^{- 31} k g$
Here $Δ m$ is $(-) v e$ so the reaction absorbs energy.The energy absorbed during the process
$E = Δ m c^{2}$
$E = 9 \times 10^{- 31} \times 9 \times 10^{16} = 81 \times 10^{- 15} joules$
$E = \frac{81 \times 10^{- 15}}{1.6 \times 10^{- 19}} = 0.511$ $M e v$
The correct answer is $- 0.511 M e V$

PHXII13:NUCLEI

363635 The difference between the mass of a nucleus and the combined mass of its nucleons is

1 Zero, positive or negative

2 Zero

3 Positive

4 Negative

PHXII13:NUCLEI

363636 $6.4 \times 10^{- 19}$ joule is approximately

1 4 electron volt

2 6 electron volt

3 8 electron volt

4 1 electron volt

Explanation:

$1 e V = 1.6 \times 10^{- 19} J$
$∴ 6.4 \times 10^{- 19} J = \frac{6.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 4 e V .$
So correct option is (1)

PHXII13:NUCLEI

363632 The energy equivalent of $0.5 g$ of a substance is:

1

4.5 \times 10^{13} J

2

1.5 \times 10^{13} J

3

0.5 \times 10^{13} J

4

3.5 \times 10^{13} J

Explanation:

$E = m c^{2}$
$= 0.5 \times 10^{- 3} \times 9 \times 10^{16}$
$= 4.5 \times 10^{13} J$

NEET - 2020

PHXII13:NUCLEI

363633 The energy supplied by a power plant is 40 million kilowatt hour. It is supplied by annihilation of matter, the mass that is annihilated is

1

1.6 g

2

1.6 k g

3

1.6 m g

4

1.6 a m u

Explanation:

$E = M C^{2}, E = 40 \times 10^{6} \times 10^{3} \times 60 \times 60 J$ .
$M = \frac{40 \times 10^{6} \times 10^{3} \times 60 \times 60}{9 \times 10^{16}}$
$= \frac{4}{9} \times 36 \times 10^{- 4} = 16 \times 10^{- 4} k g = 1.6 g .$

PHXII13:NUCLEI

363634 Assume that a neutron breaks into a proton and an electron. The energy absorbed during this process is: (mass of neutron $= 1.6725 \times 10^{- 27} k g,$ mass of proton $= 1.6725 \times 10^{- 27} k g,$ mass of electron $= 9 \times 10^{- 31} k g$ ).

1

- 0.51 M e V

2

- 7.10 M e V

3

- 6.30 M e V

4

- 5.4 M e V

Explanation:

$_{0}^{1} n \to_{1}^{1} P +_{- 1} e^{0} + \bar{v} + Q$
The mass defect during the process
$Δ m = m_{n} - m_{H} - m_{e} = 1.6725 \times 10^{27}$
$- (1.6725 \times 10^{- 27} + 9 \times 10^{- 31} k g)$
$= - 9 \times 10^{- 31} k g$
Here $Δ m$ is $(-) v e$ so the reaction absorbs energy.The energy absorbed during the process
$E = Δ m c^{2}$
$E = 9 \times 10^{- 31} \times 9 \times 10^{16} = 81 \times 10^{- 15} joules$
$E = \frac{81 \times 10^{- 15}}{1.6 \times 10^{- 19}} = 0.511$ $M e v$
The correct answer is $- 0.511 M e V$

PHXII13:NUCLEI

363635 The difference between the mass of a nucleus and the combined mass of its nucleons is

1 Zero, positive or negative

2 Zero

3 Positive

4 Negative

PHXII13:NUCLEI

363636 $6.4 \times 10^{- 19}$ joule is approximately

1 4 electron volt

2 6 electron volt

3 8 electron volt

4 1 electron volt

Explanation:

$1 e V = 1.6 \times 10^{- 19} J$
$∴ 6.4 \times 10^{- 19} J = \frac{6.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 4 e V .$
So correct option is (1)

PHXII13:NUCLEI

363632 The energy equivalent of $0.5 g$ of a substance is:

1

4.5 \times 10^{13} J

2

1.5 \times 10^{13} J

3

0.5 \times 10^{13} J

4

3.5 \times 10^{13} J

Explanation:

$E = m c^{2}$
$= 0.5 \times 10^{- 3} \times 9 \times 10^{16}$
$= 4.5 \times 10^{13} J$

NEET - 2020

PHXII13:NUCLEI

363633 The energy supplied by a power plant is 40 million kilowatt hour. It is supplied by annihilation of matter, the mass that is annihilated is

1

1.6 g

2

1.6 k g

3

1.6 m g

4

1.6 a m u

Explanation:

$E = M C^{2}, E = 40 \times 10^{6} \times 10^{3} \times 60 \times 60 J$ .
$M = \frac{40 \times 10^{6} \times 10^{3} \times 60 \times 60}{9 \times 10^{16}}$
$= \frac{4}{9} \times 36 \times 10^{- 4} = 16 \times 10^{- 4} k g = 1.6 g .$

PHXII13:NUCLEI

363634 Assume that a neutron breaks into a proton and an electron. The energy absorbed during this process is: (mass of neutron $= 1.6725 \times 10^{- 27} k g,$ mass of proton $= 1.6725 \times 10^{- 27} k g,$ mass of electron $= 9 \times 10^{- 31} k g$ ).

1

- 0.51 M e V

2

- 7.10 M e V

3

- 6.30 M e V

4

- 5.4 M e V

Explanation:

$_{0}^{1} n \to_{1}^{1} P +_{- 1} e^{0} + \bar{v} + Q$
The mass defect during the process
$Δ m = m_{n} - m_{H} - m_{e} = 1.6725 \times 10^{27}$
$- (1.6725 \times 10^{- 27} + 9 \times 10^{- 31} k g)$
$= - 9 \times 10^{- 31} k g$
Here $Δ m$ is $(-) v e$ so the reaction absorbs energy.The energy absorbed during the process
$E = Δ m c^{2}$
$E = 9 \times 10^{- 31} \times 9 \times 10^{16} = 81 \times 10^{- 15} joules$
$E = \frac{81 \times 10^{- 15}}{1.6 \times 10^{- 19}} = 0.511$ $M e v$
The correct answer is $- 0.511 M e V$

PHXII13:NUCLEI

363635 The difference between the mass of a nucleus and the combined mass of its nucleons is

1 Zero, positive or negative

2 Zero

3 Positive

4 Negative

PHXII13:NUCLEI

363636 $6.4 \times 10^{- 19}$ joule is approximately

1 4 electron volt

2 6 electron volt

3 8 electron volt

4 1 electron volt

Explanation:

$1 e V = 1.6 \times 10^{- 19} J$
$∴ 6.4 \times 10^{- 19} J = \frac{6.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 4 e V .$
So correct option is (1)