What wavelength of light will be emitted from a hydrogen atom for an electron that falls from n = 5 to n = 2 ?

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Answer 1 · 2016-08-19T17:06:08

$sf(lambda=434color(white)(x)nm)$

We can use the Rydberg Expression:

$sf(1/lambda=R[1/n_1^2-1/n_2^2])$

$sf(lambda)$ is the wavelength

$sf(R)$ is the Rydberg Constant which is $sf(1.097xx10^(-7)color(white)(x)m^-1)$

$sf(n_1)$ is the lower energy level

$sf(n_2)$ is the higher energy level

Putting in the numbers:

$sf(1/lambda=1.097xx10^7[1/2^2-1/5^2])$

$sf(1/lambda=1.097xx10^7[0.25-0.04]=0.2304xx10^7color(white)(x)m^-1)$

$:.$ $sf(lambda=4.34xx10^(-7)color(white)(x)m)$

$sf(lambda=434color(white)(x)nm)$

Transitions like this down to the n = 2 energy level form the "Balmer Series" of spectral lines.

These occur in the visible region of the electromagnetic spectrum.

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