An electron in an excited hydrogen atom makes two transitions. First the electro

Question

An electron in an excited hydrogen atom makes two transitions. First the electron drops from the n=5 to the n=3 state, then the electron drops from the n=3 to the n=1 state.

Please explain your answers

Compare the energy and wavelength of the photon emitted in each transition.

a) in the 1st transition the energy of the photon and the wavelength of the photon are both greater

            b) in the 1st transition the energy of the photon is greater and the wavelength of the photon is less

            c) in the 1st transition the energy of the photon is less and the wavelength of the photon is greater

            d) in the 1st transition the energy of the photon and the wavelength of the photon are both less

            e) none of the above

Calculate the frequency of the photon emitted in the first transition.

            a) 1.34x1014 Hz

            b) 2.34x1014 Hz

            c) 3.34x1014 Hz

            d) 4.34x1014 Hz

            e) 5.34x1014 Hz

Calculate the momentum of the photon emitted in the second transition.

a) 3.46x10-27 kg-m/s

            b) 4.46x10-27 kg-m/s

            c) 5.46x10-27 kg-m/s

            d) 6.46x10-27 kg-m/s

            e) 7.46x10-27 kg-m/s

Which of the following quantum states could the electron have been in at some point during these transitions?

a) n = 1, l = 1, m l = 0, ms = +½

            b) n = 3, l = -1, m l = 0, ms = -½

            c) n = 3, l = 0, m l = 1, ms = +½

            d) n = 5, l = 3, m l = -1, ms = +½

            e) n = 5, l = 1, m l = -3, ms = -½

Explanation / Answer

In the hydrogen atom, with Z = 1, the energy of the emitted photon is given by -
E = (13.6 eV) [1/nf2 - 1/ni2]
nf - Final State
ni  - Initial State

a)
For first transition from n = 5 to n = 3
E1 = (13.6 eV) [1/32 - 1/52]
E1 = 0.967 eV
E1 = 0.967 *  1.60 x 10-19 J = 1.547 *10-19 J

For Second transition from n = 3 to n = 1
E2 = (13.6 eV) [1/12 - 1/32]
E2 = 12.09 eV
E2 = 12.09 *  1.60 x 10-19 J = 19.34 *10-19 J

From Above E2 > E1 and we know The smaller the energy the longer the wavelength 1 > 2
Therefore -
in the 1st transition the energy of the photon is less and the wavelength of the photon is greater

b) Frequency = E/h
h = 6.63 * 10^ -34

f= 1.547 *10-19 / 6.63 * 10^ -34
Frequency (f) = 2.34 * 10^14 Hz

c)
Momentum (p) = E/C
For second transition-
p =  1.547 *10-19 / 3 * 10^8

p = 6.46 * 10^-27 kg-m/s

d)
n = 3, l = 0, m l = 1, ms = +½

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