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Standard XII

Physics

Law of Radioactivity

If the masses...

Question

If the masses of deuterium and that of helium are 2.0140 amu and 4.0026 amu, respectively and that 22.4 MeV energy is liberated in the reaction $_{6}^{3} L i +_{1}^{2} H \to_{2}^{4} H +_{2}^{4} H e$ , has the mass of $_{3}^{6} L i$ is

6.015 amu

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6.068 amu

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5.980 amu

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6.00 amu

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Solution

The correct option is A 6.015 amu
$L i_{3} + 2_{1} H \to 4_{2} H + 4_{2} H e$
$Δ m = - (2 \times 4 \cdot 0026 - 2 \cdot 0140 - m)$
$m - 5 \cdot 9912$
$22 \cdot 4 = m - 5 \cdot 9912 \times 931 \cdot 5$
$0 \cdot 024 = m - 5 \cdot 9912$
$\Rightarrow m = 6 \cdot 0152 a m u$

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Similar questions

Q. The isotopic masses of

_{1}^{2} H

and

_{2}^{4} He

are

2.0141

and

4.0026

amu respectively. The velocity of light in vacuum is

2.998 \times 10^{8} m / s e c

. The quantity of energy liberated when two mole of

_{1}^{2} H

undergo fusion to form 1 mole of

_{2}^{4} He

is :

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

amu

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. 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.

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]

Q. 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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If the masses of deuterium and that of helium are 2.0140 amu and 4.0026 amu, respectively and that 22.4 MeV energy is liberated in the reaction 36Li+21H→42H+42He, has the mass of 63Li is

The correct option is A 6.015 amuLi3+21H→42H+42HeΔm=−(2×4⋅0026−2⋅0140−m)m−5⋅991222⋅4=m−5⋅9912×931⋅50⋅024=m−5⋅9912⇒m=6⋅0152amu

If the masses of deuterium and that of helium are 2.0140 amu and 4.0026 amu, respectively and that 22.4 MeV energy is liberated in the reaction $_{6}^{3} L i +_{1}^{2} H \to_{2}^{4} H +_{2}^{4} H e$ , has the mass of $_{3}^{6} L i$ is

The correct option is A 6.015 amu
$L i_{3} + 2_{1} H \to 4_{2} H + 4_{2} H e$
$Δ m = - (2 \times 4 \cdot 0026 - 2 \cdot 0140 - m)$
$m - 5 \cdot 9912$
$22 \cdot 4 = m - 5 \cdot 9912 \times 931 \cdot 5$
$0 \cdot 024 = m - 5 \cdot 9912$
$\Rightarrow m = 6 \cdot 0152 a m u$