Two of the emission wavelengths in the hydrogen emission spectrum are 410 nm and 434 nm. One of these is due to the [tex]n=6[/tex] to [tex]n=2[/tex] transition, and one is due to the [tex]n=5[/tex] to [tex]n=2[/tex] transition. Which wavelength corresponds to which transition, and why?

The 410 mm wavelength corresponds to the n=6 to n=2 transition, the 434 mm wavelength corresponds to the n=5 to n=2 transition.
Planck's equation: E=hc/wavelength
n=6 to n=2 corresponds to a higher energy transition and hence shorter wavelength, 410nm.

Answered by priyambaksi | 2024-06-10

The wavelengths of 410 nm and 434 nm in the hydrogen emission spectrum refer to specific transitions between energy levels of the hydrogen atom. According to the given information, the n = 5 to n = 2 transition corresponds to the wavelength of 434 nm (blue), and the n = 6 to n = 2 transition corresponds to 410 nm (violet). This is due to the fact that transitions to a lower energy level result in the emission of a photon, and the energy of this photon is related to the difference in energy levels, which determines the wavelength of the emitted radiation.

Answered by KarenLucilleHale | 2024-06-18

The wavelength of 410 nm corresponds to the transition from n=6 to n=2, while 434 nm corresponds to n=5 to n=2. Shorter wavelengths signify higher energy transitions, therefore 410 nm, being shorter, represents the greater energy difference. This aligns with the principles of the Bohr model and the Balmer series for hydrogen emissions.
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Answered by priyambaksi | 2024-09-27