Assuming the splitting of `U^235` nucleus liberates 200 MeV energy, find
(a) the energy liberated in the fission of 1 kg of `U^235` and
(b) the mass of the coal with calorific value of `30 kJ//g` which is equivalent to 1 kg of `U^235`.

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Energy Liberated in the Fission of 1 kg of \( U^{235} \)

1. **Calculate the number of nuclei in 1 kg of \( U^{235} \)**:
- The molar mass of \( U^{235} \) is approximately 235 g/mol.
- The number of moles in 1 kg (1000 g) of \( U^{235} \) is:
\[ \text{Number of moles} = \frac{1000 \text{ g}}{235 \text{ g/mol}} \approx 4.2553 \text{ moles} \]
- Using Avogadro's number (\( 6.022 \times 10^{23} \) nuclei/mol), the total number of nuclei is:
\[ \text{Number of nuclei} = 4.2553 \text{ moles} \times 6.022 \times 10^{23} \text{ nuclei/mol} \approx 2.56 \times 10^{24} \text{ nuclei} \]

2. **Calculate the energy liberated**:
- Each fission of \( U^{235} \) liberates 200 MeV of energy.
- Convert MeV to Joules using the conversion factor \( 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \):
\[ \text{Energy per fission} = 200 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 3.2 \times 10^{-11} \text{ J} \]
- Total energy liberated in the fission of 1 kg of \( U^{235} \):
\[ \text{Total energy} = \text{Number of nuclei} \times \text{Energy per fission} = 2.56 \times 10^{24} \text{ nuclei} \times 3.2 \times 10^{-11} \text{ J} \approx 8.19 \times 10^{13} \text{ J} \]

### Part (b): Mass of Coal Equivalent to the Energy Liberated

1. **Given the calorific value of coal**:
- The calorific value of coal is given as \( 30 \text{ kJ/g} \) or \( 30,000 \text{ J/g} \).

2. **Calculate the mass of coal required**:
- To find the mass of coal that would provide the same energy as \( 8.19 \times 10^{13} \text{ J} \):
\[ \text{Mass of coal} = \frac{\text{Total energy}}{\text{Calorific value}} = \frac{8.19 \times 10^{13} \text{ J}}{30,000 \text{ J/g}} \approx 2.73 \times 10^{9} \text{ g} \]
- Convert grams to kilograms:
\[ \text{Mass of coal in kg} = \frac{2.73 \times 10^{9} \text{ g}}{1000} \approx 2.73 \times 10^{6} \text{ kg} \]

### Final Answers:
- (a) The energy liberated in the fission of 1 kg of \( U^{235} \) is approximately \( 8.19 \times 10^{13} \text{ J} \).
- (b) The mass of coal equivalent to this energy is approximately \( 2.73 \times 10^{6} \text{ kg} \).