Step 1: Energy Calculation using \( E = \Delta mc^2 \)
The energy \( E \) is calculated using Einstein's famous equation \( E = \Delta mc^2 \), where:
\[ E = 0.4 \times 10^{-3} \times (3 \times 10^8)^2. \]
Simplifying, we get:
\[ E = 3600 \times 10^7 \, \text{kWs}. \]
Step 2: Converting to kWh
To convert from kilowatt-seconds (kWs) to kilowatt-hours (kWh), we divide by 3600 (the number of seconds in an hour):
\[ \frac{3600 \times 10^7}{3600} \, \text{kWh} = 1 \times 10^7 \, \text{kWh}. \]
The energy liberated is given by:
\[ E = \Delta m c^2 \]
Substituting the values:
\[ E = 0.4 \times 10^{-3} \times (3 \times 10^8)^2 \]
\[ E = 3600 \times 10^7 \, \text{kWs} \]
Converting to kWh:
\[ E = \frac{3600 \times 10^7 \, \text{kWh}}{3600} = 1 \times 10^7 \, \text{kWh} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

The electric potential at the surface of an atomic nucleus \( (z = 50) \) of radius \( 9 \times 10^{-13} \) cm is \(\_\_\_\_\_\_\_ \)\(\times 10^{6} V\).
In a nuclear fission reaction of an isotope of mass \( M \), three similar daughter nuclei of the same mass are formed. The speed of a daughter nuclei in terms of mass defect \( \Delta M \) will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)