Question

A hydrogen atom

(Z = 1)

is in the fourth excited state, and a photon is either emitted or absorbed.

Concepts:
(i) What is the quantum number of the fourth excited state?

12    345

(ii) When an atom emits a photon, is the final quantum number

n_f

of the atom greater than or less than the initial quantum

number n_i?

greater thanless than    

(iii) When an atom absorbs a photon, is the final quantum number

n_f

of the atom greater than or less than the initial quantum

number n_i?

greater thanless than    

(iv) How is the wavelength of a photon related to its energy? (Use the following as necessary: c, E, and h.)
? =

Calculations:
(a) Determine the quantum number

n_f

of the final state and the energy of the photon when the photon is emitted with the shortest possible wavelength.

nf		---Select--- 1 2 3 4 5 6
Ephoton	=	eV

(b) Determine the quantum number

n_f

of the final state and the energy of the photon when the photon is emitted with the longest possible wavelength.

nf		---Select--- 1 2 3 4 5 6
Ephoton	=	eV

(c) Determine the quantum number

n_f

of the final state and the energy of the photon when the photon is absorbed with the longest possible wavelength.

nf		---Select--- 1 2 3 4 5 6
Ephoton	=	eV

Answer 1

i)
Principal quantum number corresponds to fourth excited state is n = 3

ii)
When a photon is emitted, its final quantum number is less than the initial one.

iii)
When a photon is absorbed, its final quantum number is greater than the initial one.

iv)
Wavelength, = hc/E

a)
When the photon is emitted, the final n will be less than the initial n. Shortest wavelength corresponds to highest energy. So final n = 1

b)
When a photon is emitted, n will be lowered.
Longest possible wavelength corresponds to smallest energy. ie the excited atom will de excite to its nearest lower state with n = 2

c)
When a photon is absorbed, n increases.
Since longest wavelength corresponds to smallest energy, atom will further excited to n = 4