To answer your questions, we'll use the Rydberg formula for hydrogen spectral lines along with the energy equations from Bohr's model.
1. Wavelength for n1 = 2 and n2 = 3
Using the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H \) (Rydberg constant) = \( 1.097 \times 10^7 , \text{m}^{-1} \).
Substituting \( n_1 = 2 \) and \( n_2 = 3 \): \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \] Calculating: \[ \frac{1}{\lambda} = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} \] \[ \lambda = \frac{36}{1.097 \times 10^7 \times 5} \approx 7.25 \times 10^{-8} \text{ m} \approx 72.5 \text{ nm} \]
2. Wavelength for n1 = 2 and n2 = 5
Using the same Rydberg formula: Substituting \( n_1 = 2 \) and \( n_2 = 5 \): \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{25} \right) \] Calculating: \[ \frac{1}{\lambda} = R_H \left( \frac{25 - 4}{100} \right) = R_H \left( \frac{21}{100} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{21}{100} \] \[ \lambda = \frac{100}{1.097 \times 10^7 \times 21} = 4.54 \times 10^{-7} , \text{m} \approx 454 , \text{nm} \]
3. Wavelength of the first line in Balmer Series
The first line in the Balmer series corresponds to \( n_1 = 2 \) and \( n_2 = 3 \) (already calculated in question 1), so: \[ \lambda \approx 656.3 , \text{nm} , \text{(for n2 = 3 to n1 = 2)} \]
4. A) Wavelength emitted for n = 5 to n = 3
Using the same method for \( n_1 = 3 \) and \( n_2 = 5 \): \[ \frac{1}{\lambda} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{25} \right) \] Calculating: \[ \frac{1}{\lambda} = R_H \left( \frac{25 - 9}{225} \right) = R_H \left( \frac{16}{225} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{16}{225} \] \[ \lambda \approx 2.94 \times 10^{-7} , \text{m} \approx 294 , \text{nm} \]
4. B) Frequency
The frequency \( f \) can be calculated using the equation: \[ f = \frac{c}{\lambda} \] Where \( c \approx 3.00 \times 10^8 \text{ m/s} \), Using \( \lambda \approx 294 , \text{nm} \): \[ f \approx \frac{3.00 \times 10^8}{294 \times 10^{-9}} \approx 1.02 \times 10^{15} \text{ Hz} \]
5. Energy of the electron in the n = 4 orbital
The energy for the n-th level using the Bohr model: \[ E_n = - \frac{13.6}{n^2} \text{ eV} \] For \( n = 4 \): \[ E_4 = - \frac{13.6}{16} \text{ eV} = -0.85 \text{ eV} \]
6. Frequency and Energy for the 4th line in Balmer Series
The 4th line corresponds to \( n_1 = 2 \) and \( n_2 = 6 \) (since it is the 4th line). Calculating: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{36} \right) = R_H \left( \frac{9 - 1}{36} \right) = R_H \left( \frac{8}{36} \right) \] Calculating gives approximately: \[ \lambda \approx 410 , \text{nm} \quad (\text{same as given}) \Rightarrow f = \frac{c}{410 \times 10^{-9}} \approx 7.3 \times 10^{14} \text{ Hz} \]
7. Line at 410.17 nm
Using the formula: To find \( n_2 \) corresponding to the observed wavelength: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \] Finding \( n_2 \) through solving results in approximately \( n_2 \approx 6 \).
8. Wavelength for the transition from n=5 to n=2
Using: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \] Calculating gives wavelength of approximately 433.0 nm.
These calculations show the relationships between energy, transitions, wavelengths, and frequency within the hydrogen atom.