Question 181: Atoms & Nuclei

Question

$$_{92}^{238}A 	o _{90}^{234}B + _2^4D + Q$$
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of $${ }_{92}^{238} \mathrm{~A}=238.05079 	imes 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},$$
mass of $${ }_{90}^{234} B=234 \cdot 04363 	imes 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},$$
mass of $$\left.{ }_{2}^{4} D=4 \cdot 00260 	imes 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}ight]$$

Accepted Answer

The energy released in a nuclear reaction can be determined by the mass difference between the reactants and the products, multiplied by the speed of light squared, as per Einstein's mass-energy equivalence relation, $$E=mc^2$$.
In the given nuclear reaction, the energy released is:
$$Q = \left( m_{A} - m_{B} - m_{D} ight) 	imes 931.5 \ 	ext{MeV/c}^2$$
We are given the masses of A, B, and D as follows:
$$m_{A} = 238.05079 	imes 931.5 \ 	ext{MeV/c}^2$$
$$m_{B} = 234.04363 	imes 931.5 \ 	ext{MeV/c}^2$$
$$m_{D} = 4.00260 	imes 931.5 \ 	ext{MeV/c}^2$$
Now, we can substitute these values into the equation for Q:
$$Q = \left( (238.05079 - 234.04363 - 4.00260) 	imes 931.5 ight) \ 	ext{MeV}$$
$$Q = \left( (0.00456) 	imes 931.5 ight) \ 	ext{MeV}$$
Calculating the energy released:
$$Q \approx 4.25 \ 	ext{MeV}$$
So, the approximate amount of energy released in the given nuclear reaction is 4.25 MeV.

Answer

2.12 MeV

Answer

4.25 MeV

Answer

3.82 MeV

Answer

5.9 MeV

Atoms & Nuclei

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