Which of the following transitions (in a hydrogen atom) represent emission of the shortest wavelength?

• A. n = 3 to n = 1
• B. n = 2 to n = 4
• C. n = 1 to n = 4
• D. n = 5 to n = 3
• E. n = 2 to n = 5

Answer: (A)

Verified step by step guidance

1. Understand that the emission of light in a hydrogen atom occurs when an electron transitions from a higher energy level (n) to a lower energy level.
2. Recall that the energy difference between two levels is given by the Rydberg formula: \( E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.
3. Recognize that the wavelength of the emitted light is inversely proportional to the energy difference: \( \lambda = \frac{hc}{E} \), where \( h \) is Planck's constant and \( c \) is the speed of light.
4. To find the transition with the shortest wavelength, identify the transition with the largest energy difference, as a larger energy difference corresponds to a shorter wavelength.
5. Calculate the energy differences for each transition using the Rydberg formula and compare them to determine which transition has the largest energy difference, thus the shortest wavelength.