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Mass Energy Equivalence

The atomic ma...

Question

The atomic masses of the hydrogen isotopes are
$m_{1} H^{1} = 1.007825 a m u$
$m_{1} H^{2} = 2.014102 a m u$
$m_{1} H^{3} = 3.016049 a m u$
The energy released in the reaction
$_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} +_{1} H^{1}$
is nearly:

A

1 M e V

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B

2 M e V

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C

4 M e V

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D

8 M e V

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Solution

The correct option is D $4 M e V$
Mass defect: $Δ M = 2 m_{1} H^{2} - (m_{1} H^{1} + m_{1} H^{3})$
$∴$ $Δ M = 2 (2.014202) - (1.007825 + 3.016049) = 4.33 \times 10^{- 3}$ amu
Energy released, $E = Δ M \times 931$ $M e V$
$⟹$ $E = 4.33 \times 10^{- 3} \times 931 \approx 4 M e V$

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Similar questions

Q. The energy released in the reaction

_{1} H^{2} +_{1} H^{3}

\to_{2} H e^{4} +_{0} n^{1}

- ------- -

_{0} n^{1} = 1.00867 a m u

_{1} H^{2} = 2.01410 a m u

_{1} H^{3} = 3.0160 a m u

_{2} H e^{4} = 4.0026 a m u

Q. Hydrogen exists in three isotopic forms,

_{1} H^{1}

_{1} {H^{2}}^{2}

_{1} H^{3}

known as protium, deuterium, and tritium. Why are all the isotopes neutral in nature?

Q. Consider the following reaction

^{1} H_{2} +^{1} H_{2} \to_{2} H e^{4} + Q

m (_{1} H^{2}) = 2.0141 u, m (_{2} H e^{4}) = 4.0024 u

, then the energy Q released (in MeV) in this fusion reaction is

The correct order of wavelength of hydrogen

(_{1} H^{1})

(_{1} H^{2})

(_{1} H^{3})

moving with same kinetic energy is:

Q. A star initially has

10^{40}

deuterons. It produces energy via the processes,

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} + p

_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} + n

The masses of the nuclei are as follows,

m (_{1} H^{2}) = 2.014 amu; m (_{1} H^{3}) = 3.016 amu m (p) = 1.007 amu; m (n) = 1.008 amu m (_{2} {He}^{4}) = 4.001 amu

If the average power radiated by the star is

10^{16} W

, the deuteron supply of the star is exhausted in a time period of the order of
[Take

1 amu = 931.5 \frac{MeV}{c^{2}}

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