A red line is observed at 656.3 nm in the spectrum of atomic hydrogen. determine the values of ???? for the beginning and ending

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2 years ago

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energy levels of the electron during the emission of energy that leads to this spectral line.

1 answer:

aleksklad [387] · Answer 1 · 2023-09-13T09:01:38+03:00

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

<h2>calculation :-</h2><h3>(i) Calculation of energy of photon:-</h3>

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

<h3>(ii) Beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line.</h3>

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:-

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