The shortest wavelength of hydrogen atom in Lyman series is λ. The longest wavelength in Balmer series of He+ is
For problems involving spectral series:
• Use the Rydberg formula to calculate wavelengths for hydrogen-like atoms.
• Identify the energy levels (n1, n2) and the corresponding transitions for the given series.
• Simplify using the atomic number Z and the known relationships between series.
1. Energy Levels and Wavelengths: The energy of a transition in a hydrogen-like atom is given by:
\[E = -\frac{13.6 Z^2}{n^2},\]
where \(Z\) is the atomic number and \(n\) is the principal quantum number. The wavelength \(\lambda\) is related to the energy difference between levels:
\[\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right),\]
where \(R\) is the Rydberg constant, and \(n_1\) and \(n_2\) are the lower and higher energy levels, respectively.
2. Lyman Series (Hydrogen): The shortest wavelength in the Lyman series corresponds to a transition from \(n_2 = \infty\) to \(n_1 = 1\). For hydrogen (\(Z = 1\)):
\[\frac{1}{\lambda} = R(1)^2 \left( 1 - 0 \right) \implies \lambda = \frac{1}{R}.\]
3. Balmer Series (He\(^+\)): For He\(^+\) (\(Z = 2\)), the longest wavelength in the Balmer series corresponds to a transition from \(n_2 = 3\) to \(n_1 = 2\). The wavelength is given by:
\[\frac{1}{\lambda_{\text{Balmer}}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right).\]
Simplify:
\[\frac{1}{\lambda_{\text{Balmer}}} = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{9} \right).\]
Thus:
\[\lambda_{\text{Balmer}} = \frac{9}{5}\frac{1}{\lambda}.\]
Final Answer: \(\frac{9}{5}\lambda\).
Consider the following data:
- Heat of formation of \( CO_2(g) \) = -393.5 kJ mol\(^{-1}\)
- Heat of formation of \( H_2O(l) \) = -286.0 kJ mol\(^{-1}\)
- Heat of combustion of benzene = -3267.0 kJ mol\(^{-1}\)
The heat of formation of benzene is ……… kJ mol\(^{-1}\) (Nearest integer).
Which of the following is/are correct with respect to the energy of atomic orbitals of a hydrogen atom?
(A) \( 1s<2s<2p<3d<4s \)
(B) \( 1s<2s = 2p<3s = 3p \)
(C) \( 1s<2s<2p<3s<3p \)
(D) \( 1s<2s<4s<3d \)
Choose the correct answer from the options given below:
An ideal gas undergoes a cyclic transformation starting from point A and coming back to the same point by tracing the path A→B→C→D→A as shown in the three cases below.
Choose the correct option regarding \(\Delta U\):