The binding energy of a certain nucleus is \(18 \times 10^8 \, \text{J}\). How much is the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus?
- \(0.2 \, \mu \text{g}\)
- \(20 \, \mu \text{g}\)
- \(2 \, \mu \text{g}\)
- \(10 \, \mu \text{g}\)
The Correct Option is B
Approach Solution - 1
To solve the problem of finding the difference between the total mass of all the nucleons and the nuclear mass of a given nucleus, we need to apply the concept of mass-energy equivalence. The binding energy of the nucleus is given as \( 18 \times 10^8 \, \text{J} \).
According to Einstein's mass-energy equivalence principle, the binding energy (\( E \)) is related to the mass defect (\( \Delta m \)) by the equation:
\(E = \Delta m \cdot c^2\)
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).
We can rearrange the formula to compute the mass defect (\( \Delta m \)):
\(\Delta m = \frac{E}{c^2}\)
Substituting the given values:
\(\Delta m = \frac{18 \times 10^8 \, \text{J}}{(3 \times 10^8 \, \text{m/s})^2}\)
\(\Delta m = \frac{18 \times 10^8 \, \text{J}}{9 \times 10^{16} \, \text{m}^2/\text{s}^2}\)
\(\Delta m = 2 \times 10^{-8} \, \text{kg}\)
Since the options are given in micrograms, we can convert kilograms to micrograms:
1 kilogram = \( 10^9 \) micrograms, so:
\(\Delta m = 2 \times 10^{-8} \, \text{kg} = 2 \times 10^{-8} \times 10^9 \, \mu\text{g}\)
\(\Delta m = 20 \, \mu\text{g}\)
Thus, the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus is 20 micrograms.
The correct answer is: \(20 \, \mu \text{g}\).
Approach Solution -2
Using the mass-energy equivalence \( E = \Delta m c^2 \), the mass defect \( \Delta m \) is calculated as:
\[ \Delta m = \frac{E}{c^2}. \]
Substituting the values:
\[ E = 18 \times 10^8 \, \text{J}, \, c = 3 \times 10^8 \, \text{m/s}, \]
\[ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2} = \frac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8} \, \text{kg}. \]
Converting \( \Delta m \) to micrograms (\( \mu g \)):
\[ \Delta m = 2 \times 10^{-8} \, \text{kg} = 20 \, \mu g. \]
Final Answer: 20 \( \mu g \)
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