Question

In the hydrogen atom, for which transition is the emission line observed at shorter wavelength, n=4 to n=2 or n=5 to n=1? Show the equation needed to figure this out. Consider the relative size of delta E to determine the relative wavelength.

Answer 1

In order to determine which transition in the hydrogen atom results in an emission line observed at a shorter wavelength, we can use the equation for calculating the difference in energy between two orbits.

Step-by-step explanation:

In identifying the hydrogen atom transition responsible for an emission line observed at a shorter wavelength, the Rydberg equation proves crucial.

The equation, ΔE = -Rh (1/n²f - 1/n²i), quantifies the energy difference between two orbits, where Rh is the Rydberg constant, n₂ signifies the initial orbit, and either n₄ or n₅ represents the final orbit based on the specific transition under consideration.

By computing ΔE for each transition and comparing their magnitudes, one can ascertain which transition leads to the emission line observed at the shorter wavelength.

This analytical approach enables the determination of the specific energy level changes associated with the emission spectrum of hydrogen, providing insights into the quantum nature of electron transitions within the atom.

Answer 2

The transition from n=5 to n=1 results in a shorter wavelength than the transition from n=4 to n=2 because it involves a larger difference in energy levels. We use the Rydberg formula to calculate the wavelengths and determine that larger ΔE corresponds to shorter wavelengths.

Step-by-step explanation:

Comparing Emission Line Wavelengths

To determine which transition in a hydrogen atom results in a shorter wavelength emission line, we need to consider the difference in energy levels (ΔE) between the initial and final states of the electron. To find these energy levels, we use the Rydberg formula for calculating the wavelengths of emitted or absorbed radiation:

E = hc/λ = 13.6 eV (1/nm²- 1/nn²)

where λ is the wavelength, h is Planck's constant, c is the speed of light, and E is the energy of the photon.

ΔE

is greater for transitions with larger differences in the principal quantum numbers. Therefore, comparing the transitions n=4 to n=2 and n=5 to n=1:

• For n=4 to n=2, ΔE corresponds to 13.6 eV (1/2² - 1/4²)
• For n=5 to n=1, ΔE corresponds to 13.6 eV (1/1² - 1/5²)

Since the energy difference is larger for the transition from n=5 to n=1, the wavelength for this transition will be shorter than for the n=4 to n=2 transition. In general, shorter wavelengths correspond to larger energy differences between the energy levels involved in the transition.