Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Hint:

Total Energy = Number of Nuclei $\times$ Energy per Nucleus. Remember $N = \frac{m}{M} N_A$.

Answer 1

Correct Answer: 228

The problem requires calculating the energy released when all atoms in 47 g of $^{235}_{92}U$ undergo fission, with an average energy release of 190 MeV per fission. We need the energy in terms of $\alpha$ such that it equals $\alpha \times 10^{23}$ MeV. Let's begin by determining the number of atoms in 47 g of $^{235}_{92}U$.

Step 1: Calculate the number of moles of $^{235}_{92}U$
The molar mass of $^{235}_{92}U$ is 235 g/mol. Thus, the number of moles ($n$) in 47 g is:

$n = \frac{47 \text{ g}}{235 \text{ g/mol}} = 0.2 \text{ mol}$

Step 2: Calculate the number of atoms
Using Avogadro's number ($6 \times 10^{23}$ atoms/mol):

Number of atoms = $0.2 \text{ mol} \times 6 \times 10^{23} \text{ atoms/mol} = 1.2 \times 10^{23}$ atoms

Step 3: Calculate total energy released
Each fission releases 190 MeV, so the total energy is:

Total energy = $1.2 \times 10^{23} \text{ atoms} \times 190 \text{ MeV/atom} = 228 \times 10^{23}$ MeV

Thus, $\alpha = 228$.

Step 4: Validation
The computed value of $\alpha$ is 228, which falls within the given range of 228 to 228.

Therefore, the value of $\alpha$ is 228.