A common example of alpha decay is ${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2} \mathrm{He}^{4}+\mathrm{Q}$ Given : ${ }_{92}^{238} \mathrm{U}=238.05060 \mathrm{u}$, ${ }_{90}^{234} \mathrm{Th}=234.04360 \mathrm{u}$, ${ }_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}$, and $1 \mathrm{u}=931.5 \frac{\mathrm{MeV}}{\mathrm{c}^{2}}$ The energy released (Q) during the alpha decay of ${ }_{92}^{238} \mathrm{U}$ is \_\_\_\_ MeV

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Modern Physics

A common example of alpha decay is

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

Given :

_{92}^{238} U = 238.05060 u

_{90}^{234} Th = 234.04360 u

_{2}^{4} He = 4.00260 u

, and

1 u = 931.5 \frac{MeV}{c ^{2}}

The energy released (Q) during the alpha decay of

_{92}^{238} U

is ____ MeV

Answer: 4

Solution

The energy released equals the mass defect times

c^{2}

. The mass defect is

Δ m = m_{U} - m_{T h} - m_{H e} = 238.05060 - 234.04360 - 4.00260 = 0.00440 u

. Using

1 u = 931.5 MeV / c^{2}

Q = 0.00440 \times 931.5 = 4.0986 MeV

. The energy released in the alpha decay is about

4 MeV

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer

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