Question

A red line is observed at 656.3 nm in the spectrum of atomic hydrogen. determine the values of ???? for the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line.

Answer 1

The beginning and ending energy levels of the electron during the emission of energy that leads to spectral line n(i)= 2 and n(f)=2

calculation :-

(i) Calculation of energy of photon:-

wavelength(λ) = 656.3 nm = 656.3 x 10^-9

E= hc / λ

E = energy of the photon

h=Planck's constant

c=speed of light

λ= wavelength of the photon

E= (6.626 x 10^-34)(3 x 10^8)/( 656.3 x 10^-9)

E = 3.028 x 10^-37J

(ii) Beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line.

It's a red line which means it falls in a visible region which is known as the Balmer series so;

n (f) = 2

ΔE= RH (1/n(f)^2 - 1/n(i)^2)

3.028 x 10^-37 =-2.18 x 10^-18(1/2^2 - 1/n(i))

n(i)= 2

Learn more about energy of the photon here:-

brainly.com/question/15946945

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