Question

The electron in a hydrogen atom starts in the n = 7 level. Determine all the possible wavelengths that could be emitted if the electron ends in the first excited state.

Answer 1

use the formula 1/L = R *(1/n^2 -1/m^2)

here L is wavelength

R is Rydberg Constant = 1.1 *10^7 m

so

when m = 7 and n = 6

1/L1 = 1.1 *10^7 *(1/6^2-1/7^2)

L1 = 1/(1.1 *10^7 *(1/6^2-1/7^2))

L1 = 12335.6 nm

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when m = 7 and n = 5

1/L2 = 1.1 *10^7 *(1/5^2-1/7^2)

L2 = 1/(1.1 *10^7 *(1/5^2-1/7^2))

L2 = 4640.15 nm

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when m = 7 and n = 4

1/L3 = 1.1 *10^7 *(1/4^2-1/7^2)

L3 = 1/(1.1 *10^7 *(1/4^2-1/7^2))

L3 = 2159.7 nm

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when m = 7 and n = 3

1/L4 = 1.1 *10^7 *(1/3^2-1/7^2)

L4 = 1/(1.1 *10^7 *(1/3^2-1/7^2))

L4 = 1002.2nm

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when m = 7 and n = 2

1/L5 = 1.1 *10^7 *(1/2^2-1/7^2)

L5 = 1/(1.1 *10^7 *(1/2^2-1/7^2))

L5 = 396 nm

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when m = 7 and n = 1

1/L6 = 1.1 *10^7 *(1/1^2-1/7^2)

L6 = 1/(1.1 *10^7 *(1/1^2-1/7^2))

L6 = 92.8 nm

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