Calculate the Q-values of the following fusion reactions-a 12 H -12 H -13 H -11 Hb 12 H -12 H -32 He - nc 12 H -13 H -42 He - nAtomic masses are m 12 He -2-014102 u - m 13 He -3-016049 u - m 23 He -3-016029 u - m 24 He -4-002603 u-

(a) Q = [2 × m^2_1H m^3_1H + m^1_1H ] C^2 = (4.028204 4.023874) 931 MeV = 4.03 MeV (b) Q = [2 × m^2_1 H (m^3_2 H + n)] C^2 = [4.028204 4.024694) × 231 = ...

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Byju's Answer

Standard XII

Physics

Nuclear Fission

Calculate the...

Question

Calculate the Q-values of the following fusion reactions:

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

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

(c) $_{1}^{2} H +_{1}^{3} H \to_{4}^{2} H e + n$

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

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Solution

(a) Q = $[2 \times m_{1}^{2} H - m_{1}^{3} H + m_{1}^{1} H] C^{2}$

= (4.028204-4.023874) 931 MeV

= 4.03 MeV

(b) Q = $[2 \times m_{1}^{2} H - (m_{2}^{3} H + n)] C^{2}$

= [4.028204 - 4.024694) $\times 231$

= $0.0351 \times 931$

= 3.26 MeV

(c) Q = [ $m_{1}^{2} H + m_{1}^{3} H - m_{1}^{4} H e - m n]$ 931

= (2.014102 + 3.016049- 4.002603-1.008665) 931

= 17.58 Mev

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Similar questions

Q. 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 {}_{2}^{3}{He} + n

(c)

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

.
Atomic masses are

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

Q. Calculate the energy which can be obtained from

1

kg of

H_{2} O

through the fusion reaction

_{1}^{2} H

_{1}^{2} H

\to

_{1}^{3} H

p

.
Assume

1.5 \times 10^{2}

of the water contains deuterium. The whole deuterium is consumed in the fusion reaction

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

X
The product X in the nuclear reaction represented is.

Q. What is the difference between the number of neutrons of

_{1}^{2} H

and

_{1}^{3} H

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.

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Calculate the Q-values of the following fusion reactions: (a) 21H+21H→31H+11H (b) 21H+21H→23He+n (c) 21H+31H→24He+n Atomic masses are m(21He)=2.014102 u,m(31He)=3.016049 u,m(32He)=3.016029 u,m(42He)=4.002603 u.

(a) Q = [2×m21H−m31H+m11H]C2 = (4.028204-4.023874) 931 MeV = 4.03 MeV (b) Q = [2×m21 H−(m32 H+n)]C2 = [4.028204 - 4.024694) ×231 = 0.0351×931 = 3.26 MeV (c) Q = [m21 H+m31 H−m41 He−mn] 931 = (2.014102 + 3.016049- 4.002603-1.008665) 931 = 17.58 Mev