A nuclear reaction along with the masses of the particles taking part in it is as follows- A-B- C-D-QMeV amu 1-002 - amu 1-004 - amu 1-001 - amu 1-003 -QThe energy Q liberated in the reaction is

The correct option is B 1.862MeV by relation Q=∑ m _ r ∑ m _ p =2.006 2.004 ∴ Q=0.002× 931=1.86MeV ...

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Alpha Decay

A nuclear rea...

Question

A nuclear reaction along with the masses of the particles taking part in it is as follows:
$A + B ⟶ C + D + Q M e V$
$1.002 a m u + 1.004 a m u ⟶ 1.001 a m u + 1.003 a m u + Q$
The energy $Q$ liberated in the reaction is

1.234 M e V

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1.862 M e V

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0.931 M e V

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0.800 M e V

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Solution

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Similar questions

A + B

\to

C + D + Q

A

1.002

B

1.004

C

1.001

D

1.003

_{7} N^{14} +

_{2} H e^{4} \to_{8} O^{17} +

_{1} H^{1}

α

M_{N} =

14.00307

M_{H e} =

4.00260

M_{o} =

16.99914

M_{H} =

1.00783

^{14} N (d, p)^{15} N, Q = 8.53 M e V

^{15} N (d, α)^{13} C, Q = 7.58 M e V

^{13} C (d, α)^{11} B, Q = 5.16 M e V

^{14} N (d, p)^{15} N

^{14} N + d \to^{15} N + p

_{2}^{4} H e = 4.0026 a m u,_{1}^{2} H e = 2.014 a m u,_{1}^{1} H = 1.0078 a m u, n = 1.0087 a m u (1 a m u = 931 M e V)

^{11} B (α, n)^{14} N

A + B \to C

\frac{5.3}{p} M e V

= 930 M e V / c^{2} .

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