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Alpha Decay

Consider the ...

Question

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 _ $M e V$ .

Fission,

24

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Fusion,

24

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Fusion,

34

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Fusion,

44

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Solution

The correct option is D Fusion, $24$ .
Energy released
$= (2 \times 2.014 - 4.0024) \times 931.5$ Mev
$= 24$ Mev
Since two atoms combine to give one atom it is a fusion reaction.

Suggest Corrections

Similar questions

^{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

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

amu

2.015, 3.017

1.009

1 kg

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

1

H_{2} O

_{1}^{2} H

_{1}^{2} H

\to

_{1}^{3} H

p

1.5 \times 10^{2}

Calculate the Q-values of the following fusion reactions:

(a) $_{1}^{2} H +_{1}^{2} H \to_{1}^{3} H +_{1}^{1} H$

(b) $_{1}^{2} H +_{1}^{2} H \to_{3}^{2} H e + n$

Atomic masses are $m (_{1}^{2} H e) = 2.014102 u, m (_{1}^{3} H e) = 3.016049 u, m (_{2}^{3} H e) = 3.016029 u, m (_{2}^{4} H e) = 4.002603 u$ .

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

2.015, 3.017

1.009

1 k g

(1 a m u = 931.5 m e V / c^{2})

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