Calculate the energy released in MeV in the following nuclear reaction - 12H-12H- 23He-01nAssume that the masses of 12H- 23He and neutron n respectively are 2-020- 3-0160 and 1-0087 in amu-

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Standard VII

Chemistry

Sub-Atomic Particles

Calculate the...

Question

Calculate the energy released in MeV in the following nuclear reaction :
$_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n$
Assume that the masses of $_{1}^{2} H,_{2}^{3} He$ and neutron (n) respectively are 2.020, 3.0160 and 1.0087 in amu.

14.25

MeV

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1.425

MeV

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142.5

MeV

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1425

MeV

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Solution

The correct option is A $14.25$ MeV
The mass defect is the difference between the mass of reactants and the mass of products.
$Δ m = 2 \times m {_{1}}^{2} H - m {_{2}}^{3} H e - m_{n} = (2 \times 2.020) - (3.0160 + 1.0087) = 4.040 - 4.0247 = 0.0153$
The energy released during the nuclear reaction is
$Δ E = Δ m \times 931.48 M e V = 14.25 M e V$

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Similar questions

Q. In fusion reaction

_{1}^{2} H +_{1}^{2} H \to_{2}^{3} H e +_{0}^{1} n,

the masses of deuteron, helium and neutron expressed in

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are

2.015, 3.017

and

1.009

respectively. If

1 kg

of deuterium undergoes complete fusion, find the amount of total energy released

(1 amu c^{2} = 931.5 MeV)

Q. Consider the following reaction

_{1}^{2} H +_{1}^{2} H =_{2}^{4} H e + Q

Mass of the deuterium atom

= 2.0141 u

Mass of helium atom

= 4.0024 u

This is a nuclear ________ reaction in which the energy

Q

released is _______

M e V

Q. In the reaction

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n

. If the binding energies of

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

and

_{2}^{4} He

are respectively

a, b

and

c

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

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV]

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_{1}^{2} H e +_{1}^{2} H \to_{2}^{3} H e +_{0}^{1} n

, the masses of deuteron, helium, and neutron expressed in amu are

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Calculate the energy released in MeV in the following nuclear reaction :21H+21H→32He+10nAssume that the masses of 21H, 32He and neutron (n) respectively are 2.020, 3.0160 and 1.0087 in amu.

The correct option is A 14.25 MeVThe mass defect is the difference between the mass of reactants and the mass of products.Δm=2×m 12H−m 23He−mn=(2×2.020)−(3.0160+1.0087)=4.040−4.0247=0.0153The energy released during the nuclear reaction is ΔE=Δm×931.48MeV=14.25MeV

Calculate the energy released in MeV in the following nuclear reaction :
$_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n$
Assume that the masses of $_{1}^{2} H,_{2}^{3} He$ and neutron (n) respectively are 2.020, 3.0160 and 1.0087 in amu.