thank you Consider a nuclear reaction32He32 He right arrow4

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Consider a nuclear reaction:^3_2He+^3_2 He right arrow^4_2He+^1_1H + X (1.0 piont) What is the element X? (0.5 pionts) The masses of^1_1H,^2_1H,^14_7N,^12_6C,^3_2He,^4_2He,^7_3Li, and^9_4Be, are 1.007825 u, 2.014102 u, 14.003074 u, 12.000000 u, 3.0160293 u, 4.002603 u, 6.941 u and 9.012182 u, respectively. What is the Q-value of this reaction? Note that 1 u = 931.494 MeV.

Solution

On left side :

3 He 2 - > 2 p and (3-2) = 1 n

3 He 2 - > 2 p and 1 n

on right side :

4 He 2 - > 2 p and 2 n

1 H 1 -> 1 p and 0 n

equation will balanced. balancing eqution.

right side will have, p = ( 2 + 2) - (2 + 1) = 1

n = (1 + 1 ) - (2 + 0) = 0

hence X is proton. .......Ans (A)

B)

total mass on left side = 2 ( 3.0160293) = 6.0320586 u

on right side = 4.002603 + 1.007825 + 1.00727647 = 6.01777045 u

mass diff = 6.0320586 - 6.01777045 = 0.014354 u

Q = 0.014354 x 931.494 = 13.37 MeV .....Ans