Energy of first Balmer line of H-atom is \( x \) kJ. The energy of the second Balmer line of H-atom is _____
Step 1: Energy of the Balmer series.
The energy of each line in the Balmer series for hydrogen can be given by the Rydberg formula. The energy for the first line (n=3 to n=2) and the second line (n=4 to n=2) are related.
Step 2: Energy ratio.
The energy of the second Balmer line will be higher than the first. The ratio of the energies for the first and second lines can be calculated based on the Rydberg equation, which gives approximately a factor of 1.35 times.
Step 3: Conclusion.
Thus, the energy of the second Balmer line is 1.35 times that of the first.
Final Answer: \[ \boxed{1.35x}. \]
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 : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are

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 : \(A_2 \rightleftharpoons 2A\)