Calculate the energy released in the nuclear reaction ^ 223 Ra → ^ 209 Pb + ^ 14 C . The necessary atomic masses are provided below. ^ 223 Ra ^ 209 Pb ^ 14 C 223.018 u 208.981 u 14.003 u

Correct answer: B. 31.65 MeV The mass defect Δ m in the given nuclear reaction is calculated as: Δ m = M(^ 223 Ra) - [M(^ 209 Pb) + M(^ 14 C)] Substituting the given values: Δ m = 223.018 - (208.981 + 14.003) Δ m = 223.018 - 222.984 = 0.034 u The energy released Q is given by: Q = Δ m × 931 MeV/u Q = 0.034 × 931 MeV = 31.654 MeV This is approximately 31.65 MeV .

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Nucleus

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus

Calculate the energy released in the nuclear reaction ^ 223 Ra → ^...

Calculate the energy released in the nuclear reaction ^{223}\text{Ra} \to ^{209}\text{Pb} + ^{14}\text{C}

. The necessary atomic masses are provided below.<br><table style="border: 1px solid black; border-collapse: collapse; width: auto;"><tr><td style="border: 1px solid black; padding: 10px;">^{223}Ra

^{209}Pb

</td><td style="border: 1px solid black; padding: 10px;">^{14}C

223.018 u

208.981 u

14.003 u

Options

A $28.50 \text{ MeV}$
B $31.65 \text{ MeV}$
C $15.83 \text{ MeV}$
D $63.30 \text{ MeV}$

Correct answer

B. $31.65 \text{ MeV}$

Step-by-step solution

The mass defect

\Delta m

in the given nuclear reaction is calculated as:

\Delta m = M(^{223}\text{Ra}) - [M(^{209}\text{Pb}) + M(^{14}\text{C})]

Substituting the given values:

\Delta m = 223.018 - (208.981 + 14.003)

\Delta m = 223.018 - 222.984 = 0.034 \text{ u}

The energy released

Q

is given by:

Q = \Delta m \times 931 \text{ MeV/u}

Q = 0.034 \times 931 \text{ MeV} = 31.654 \text{ MeV}

This is approximately

31.65 \text{ MeV}

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