객관식
Which equation relates the energy of a photon to its frequency?
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9. Quantum Mechanics
Wavelength and Frequency
객관식
Using the Balmer formula for hydrogen, what is the wavelength (in nm) of the spectral line produced when an electron transitions from n = 8 to n = 2?
A
434 nm
B
388 nm
C
656 nm
D
486 nm
검증된 단계별 안내1
Identify the Balmer formula for the wavelength of light emitted during an electron transition in a hydrogen atom: \(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\), where \(\lambda\) is the wavelength, \(R\) is the Rydberg constant (\(1.097 \times 10^7 \ \text{m}^{-1}\)), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.
Assign the given values: the electron transitions from \(n_2 = 8\) to \(n_1 = 2\) (since the electron moves to a lower energy level, \(n_1 < n_2\)).
Substitute the values into the formula: \(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{8^2} \right)\).
Calculate the difference inside the parentheses: \(\frac{1}{4} - \frac{1}{64}\), which simplifies the expression for \(\frac{1}{\lambda}\).
Take the reciprocal of \(\frac{1}{\lambda}\) to find \(\lambda\) in meters, then convert the wavelength to nanometers by multiplying by \$10^9$ (since \(1 \ \text{m} = 10^9 \ \text{nm}\)).
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