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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