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

1 Answer
Aug 19, 2016

Answer:

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

Explanation:

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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