Open Question
Complete this sentence: atoms emit visible and ultraviolet light
9. Quantum Mechanics
Emission Spectrum
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
When an electron in a hydrogen atom moves from n = 8 to n = 2, light with what wavelength is emitted?
A
657 nm
B
93.8 nm
C
411 nm
D
487 nm
E
389 nm
Verified step by step guidance1
Identify the initial and final energy levels of the electron transition: n_initial = 8 and n_final = 2.
Use the Rydberg formula to calculate the wavelength of the emitted light: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \), where \( R_H \) is the Rydberg constant for hydrogen (1.097 x 10^7 m^-1).
Substitute the values for n_initial and n_final into the Rydberg formula: \( \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{8^2} \right) \).
Calculate the difference in the fractions: \( \frac{1}{4} - \frac{1}{64} \), and simplify the expression.
Solve for \( \lambda \) by taking the reciprocal of the result from the Rydberg formula calculation to find the wavelength of the emitted light.
Related Videos
Related Practice
Emission Spectrum practice set
Problem sets built by lead tutorsExpert video explanations