Question

For the given radioactive decay \(^{298}_{94}X → ^{294}_{92}X + ^{4}_{2}ɑ + Q\) - value, binding energy per nucleon of X, Y and ɑ are a, b and c. The Q - value is equal to

• A. 294b + 4c - 298a
• B. 92b + 2c - 94a
• C. 92b + 2c - 94a
• D. 92b + 2c + 94a

Answer 1

The Correct Option is A

Sure! Here is the solution in a CKEditor-compatible HTML format:

To find the Q-value of the given nuclear reaction, we need to understand the concept of binding energy and how it relates to the energy released or absorbed in a nuclear reaction.

The given reaction is:

\(^{298}_{94}X → ^{294}_{92}Y + ^{4}_{2}ɑ + Q\)

In a nuclear reaction, the Q-value is the difference between the total initial binding energy and the total final binding energy. It is defined as:

\(Q = (\text{Initial binding energy}) - (\text{Final binding energy})\)

The binding energy for a nucleus is given as the product of the binding energy per nucleon and the number of nucleons. Therefore:

• Binding energy of \(^{298}_{94}X\) = \(298a\)
• Binding energy of \(^{294}_{92}Y\) = \(294b\)
• Binding energy of \(^{4}_{2}ɑ\) = \(4c\)

Thus, the Q-value of the reaction is:

\(Q = (294b + 4c) - 298a\)

This simplifies to:

\(Q = 294b + 4c - 298a\)

Therefore, the correct answer is 294b + 4c - 298a.