Question

Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron fallsfrom the n = 7 to the n = 4 principal energy level. (RH = 2.18 x 10-18J, 1 nm = 1 x 10-9 m, h = 6.63 x 10-34 J·s,and c = 3.00 x 108 m/s)

Answer 1

Answer:

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

Explanation:

We recall that Eₙ=$\frac{-2.18*10^{-18} J}{n^{2} }$

since there was transition from n7 to n=4 we will first calculate the change in the energy i.e ΔE

ΔE=E₄-E₇

ΔE=$-2.18*10^{-18} J(\frac{1}{4^{2} } -\frac{1}{7^{2} } )$

ΔE=-9.1760*10^-20 J

Now:

|ΔE|=Energy of photon=h*v=h*c/λ

λ=h*c/|ΔE|

λ=$\frac{6.63*10^{-34}J.s*3.00*10^{8}m/s }{9.1760*10^{-20} J }$

λ=2.1676*10^-6 m

λ=2167.6*10^-9 m

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.