The Lyman series results from transitions of the electron in hydrogen in which t

Question

The Lyman series results from transitions of the electron in hydrogen in which the electrons ends at the n equal to 1 energy level. Using the Rydberg formula for the Lyman series, calculate the wavelengths of the photons emitted in the transitions from the energy states that correspond to n equal to 2 through 6, and indicate the initial and final state of the transition corresponding to each wavelength.

from m = 2 to n = 1 _______nm

from m = 3 to n = 1 ________ nm

from m = 4 to n = 1 _________nm

from m = 5 to n = 1 _________nm

from m = 6 to n = 1 __________nm

Explanation / Answer

Rydberg formula:

1/ = R(1/n2 - 1/m2)

where R = 1.097 * 107 m-1

For Lyman series, n = 1

So, 1/ = R(1 - 1/m2)

=> = 1 / [R(1 - 1/m2)]

For m = 2 to n = 1,

= 1 / [(1.097 * 107) * (1 - 1/22)] = 1.215 * 10-7 m = 121.5 nm

For m = 3 to n = 1,

= 1 / [(1.097 * 107) * (1 - 1/32)] = 1.025 * 10-7 m = 102.5 nm

For m = 4 to n = 1,

= 1 / [(1.097 * 107) * (1 - 1/42)] = 0.9723 * 10-7 m = 97.23 nm

For m = 5 to n = 1,

= 1 / [(1.097 * 107) * (1 - 1/52)] = 0.9496 * 10-7 m = 94.96 nm

For m = 6 to n = 1,

= 1 / [(1.097 * 107) * (1 - 1/62)] = 0.9376 * 10-7 m = 93.76 nm

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