The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^(7)Li+._(1)^(1)H rarr ._(2)^(4)He+._(2)^(4)He+Q`, the value of energy `Q` released is

Options

A. `19.6` MeV
B. `-2.4` MeV
C. `8.4` MeV
D. `17.3` MeV

Answer

To find the energy \( Q \) released in the nuclear reaction \( _{3}^{7}Li + _{1}^{1}H \rightarrow _{2}^{4}He + _{2}^{4}He + Q \), we will follow these steps:

### Step 1: Calculate the Binding Energy of Lithium-7
The binding energy per nucleon of Lithium-7 (\( _{3}^{7}Li \)) is given as \( 5.60 \) MeV. Since the mass number \( A \) of Lithium-7 is \( 7 \), we can calculate the total binding energy \( BE_{Li} \) as follows:

\[ BE_{Li} = A \times \text{(Binding Energy per Nucleon)} = 7 \times 5.60 \, \text{MeV} = 39.2 \, \text{MeV} \]

### Step 2: Calculate the Binding Energy of Helium-4
The binding energy per nucleon of Helium-4 (\( _{2}^{4}He \)) is given as \( 7.06 \) MeV. Since the mass number \( A \) of Helium-4 is \( 4 \), we can calculate the total binding energy \( BE_{He} \) for two Helium-4 nuclei as follows:

\[ BE_{He} = 2 \times (A \times \text{(Binding Energy per Nucleon)}) = 2 \times (4 \times 7.06 \, \text{MeV}) = 2 \times 28.24 \, \text{MeV} = 56.48 \, \text{MeV} \]

### Step 3: Assume the Binding Energy of Hydrogen
The binding energy of a hydrogen nucleus (\( _{1}^{1}H \)) is approximately \( 0 \) MeV since it has no binding energy (it is a single nucleon).

### Step 4: Calculate the Energy Released \( Q \)
The energy released \( Q \) in the reaction can be calculated using the difference in binding energies:

\[ Q = BE_{He} - BE_{Li} - BE_{H} \]

Substituting the values we calculated:

\[ Q = 56.48 \, \text{MeV} - 39.2 \, \text{MeV} - 0 \, \text{MeV} = 17.28 \, \text{MeV} \]

### Step 5: Round the Result
Rounding \( 17.28 \, \text{MeV} \) gives approximately \( 17.3 \, \text{MeV} \).

Thus, the energy \( Q \) released in the reaction is:

\[ \boxed{17.3 \, \text{MeV}} \]