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A ground state hydrogen atom absorbs a photon of light having a wavelength of 92

ID: 475583 • Letter: A

Question


A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.57 nm. It then gives off a photon having a wavelength of 1944nm. What is the final state of the hydrogen atom
A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.57 nm. It then gives off a photon having a wavelength of 1944nm. What is the final state of the hydrogen atom
A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.57 nm. It then gives off a photon having a wavelength of 1944nm. What is the final state of the hydrogen atom

Explanation / Answer

Expression of energy (E) and wavelength () can be given as follows:

E = hc/

Where,

h = Plank’s constant = 6.626 x 10-34 J.s

c = velocity of light = 3.0 x 108 m/s

Thus, Energy absorbed by ground state electron (Ea) and energy emitted by excited state electron (Ee) can be written as follows:

Ea = hc / (92.57 nm)

Ee = hc / (1944 nm)

Energy difference between Ea and Ee can be calculated as follows:

E = Ea – Ee

E = [hc / (92.57 nm)] – [ hc / (1944 nm)]

E = hc [(1 / 92.57 x 10-9 m) – (1 / 1944 x 10-9 m)]

E = (6.626 x 10-34 J.s) x (3.0 x 108 m/s) (0.01080 x 10+9 – 0.0005144 x 10+9)m-1

E = (6.626 x 10-34 J.s) x (3.0 x 108 m/s) (0.01029 x 10+9m-1)

E =0.2045 x 10-17 J

Thus, energy of final state(Ef) can be calculated as follows:

Ef = Energy of ground state + E

Ef = -(21.79 x 10-19 J) + (0.2045 x 10-17 J)

Ef = -1.34 x 10-19 J

Expression for energy (Ef) of nth energy state in hydrogen atom can be written as follows:

Ef = -Eo/n2

Where, Eo = energy of ground state = 21.79 x 10-19 J

Thus, Ef = - (21.79 x 10-19 J) / n2

n2 = - (21.79 x 10-19 J) / Ef

Substitute Ef = -1.34 x 10-19 J

n2 = - (21.79 x 10-19 J) / (-1.34 x 10-19 J)

n2 = 16.26

n = 4.0

Therefore, final state of the hydrogen atom is n = 4.

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