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CBSE Class 12 · 2008 · All India · Modern PhysicsShort Answer

Calculate the energy released in MeV in the following nuclear reaction:

_{92}^{238} U \to_{90}^{234} Th +_{2}^{4} He + Q

. [Mass of

_{92}^{238} U = 238.05079 u

; mass of

_{90}^{234} Th = 234.043630 u

; mass of

_{2}^{4} He = 4.002600 u

;

1 u = 931.5 MeV / c^{2}

.]

Answer: $Q \approx 4.25 MeV$

Solution

Mass defect:

Δ m = 238.05079 - 234.043630 - 4.002600 = 0.004560 u .

Energy released:

Q = Δ m \times 931.5 = 0.004560 \times 931.5 \approx 4.25 MeV .

Q \approx 4.25 MeV

CBSE Class 12 Board · 2008 · All India · Subjective

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