Answer :
Sure, let's go through the problem step-by-step.
### Given:
- The Rydberg constant, [tex]\( R_H = 2.18 \times 10^{-18} \, \text{J} \)[/tex]
- Speed of light, [tex]\( c = 3.00 \times 10^8 \, \text{m/s} \)[/tex]
- Planck's constant, [tex]\( h = 6.63 \times 10^{-34} \, \text{J}\cdot\text{s} \)[/tex]
- Avogadro's number, [tex]\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)[/tex]
### Problem:
Calculate for an electron in the hydrogen atom which falls from [tex]\( n = 5 \)[/tex] to [tex]\( n = 3 \)[/tex]:
(a) The energy of the electron at the excited state ([tex]\( n = 5 \)[/tex]).
(b) The energy emitted by one mole of electrons.
(c) The wavelength of the emitted photon.
### Solution:
#### (a) Energy of the electron at the excited state
The energy of an electron in a hydrogen atom at a given energy level [tex]\( n \)[/tex] is given by:
[tex]\[ E_n = - R_H \left( \frac{1}{n^2} \right) \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ E_5 = - R_H \left( \frac{1}{5^2} \right) \][/tex]
[tex]\[ E_5 = - R_H \left( \frac{1}{25} \right) \][/tex]
[tex]\[ E_5 = - 2.18 \times 10^{-18} \times \frac{1}{25} \][/tex]
[tex]\[ E_5 \approx -8.72 \times 10^{-20} \, \text{J} \][/tex]
So, the energy of the electron at [tex]\( n = 5 \)[/tex] is [tex]\( -8.72 \times 10^{-20} \, \text{J} \)[/tex].
#### (b) Energy emitted by one mole of electrons
We first need to calculate the energy of the electron at the final state [tex]\( n = 3 \)[/tex].
For [tex]\( n = 3 \)[/tex]:
[tex]\[ E_3 = - R_H \left( \frac{1}{3^2} \right) \][/tex]
[tex]\[ E_3 = - R_H \left( \frac{1}{9} \right) \][/tex]
[tex]\[ E_3 = - 2.18 \times 10^{-18} \times \frac{1}{9} \][/tex]
[tex]\[ E_3 \approx -2.422222222222222 \times 10^{-19} \, \text{J} \][/tex]
The energy emitted by the electron as it transitions from [tex]\( n = 5 \)[/tex] to [tex]\( n = 3 \)[/tex]:
[tex]\[ \Delta E = E_3 - E_5 \][/tex]
[tex]\[ \Delta E = -2.422222222222222 \times 10^{-19} - (-8.72 \times 10^{-20}) \][/tex]
[tex]\[ \Delta E \approx -1.550222222222222 \times 10^{-19} \, \text{J} \][/tex]
The energy emitted by one mole of electrons:
[tex]\[ \text{Energy}_\text{mole} = \Delta E \times N_A \][/tex]
[tex]\[ \text{Energy}_\text{mole} = -1.550222222222222 \times 10^{-19} \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{Energy}_\text{mole} \approx -93354.38222222221 \, \text{J/mol} \][/tex]
So, the energy emitted by one mole of electrons is approximately [tex]\(-93354.38222222221 \, \text{J/mol}\)[/tex].
#### (c) Wavelength of the emitted photon
Using the relationship between the energy of a photon, its wavelength, and the constants [tex]\( h \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ \Delta E = h \nu \][/tex]
[tex]\[ \Delta E = \frac{h c}{\lambda} \][/tex]
Rearranging for the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{h c}{\Delta E} \][/tex]
Substituting the values:
[tex]\[ \lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.550222222222222 \times 10^{-19}} \][/tex]
[tex]\[ \lambda \approx 1.2830418577981652 \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ \lambda \approx 1283 \, \text{nm} \][/tex] (convert meters to nanometers by multiplying by [tex]\(10^9\)[/tex])
So, the wavelength of the emitted photon is approximately [tex]\( 1283 \, \text{nm} \)[/tex].
### Given:
- The Rydberg constant, [tex]\( R_H = 2.18 \times 10^{-18} \, \text{J} \)[/tex]
- Speed of light, [tex]\( c = 3.00 \times 10^8 \, \text{m/s} \)[/tex]
- Planck's constant, [tex]\( h = 6.63 \times 10^{-34} \, \text{J}\cdot\text{s} \)[/tex]
- Avogadro's number, [tex]\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)[/tex]
### Problem:
Calculate for an electron in the hydrogen atom which falls from [tex]\( n = 5 \)[/tex] to [tex]\( n = 3 \)[/tex]:
(a) The energy of the electron at the excited state ([tex]\( n = 5 \)[/tex]).
(b) The energy emitted by one mole of electrons.
(c) The wavelength of the emitted photon.
### Solution:
#### (a) Energy of the electron at the excited state
The energy of an electron in a hydrogen atom at a given energy level [tex]\( n \)[/tex] is given by:
[tex]\[ E_n = - R_H \left( \frac{1}{n^2} \right) \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ E_5 = - R_H \left( \frac{1}{5^2} \right) \][/tex]
[tex]\[ E_5 = - R_H \left( \frac{1}{25} \right) \][/tex]
[tex]\[ E_5 = - 2.18 \times 10^{-18} \times \frac{1}{25} \][/tex]
[tex]\[ E_5 \approx -8.72 \times 10^{-20} \, \text{J} \][/tex]
So, the energy of the electron at [tex]\( n = 5 \)[/tex] is [tex]\( -8.72 \times 10^{-20} \, \text{J} \)[/tex].
#### (b) Energy emitted by one mole of electrons
We first need to calculate the energy of the electron at the final state [tex]\( n = 3 \)[/tex].
For [tex]\( n = 3 \)[/tex]:
[tex]\[ E_3 = - R_H \left( \frac{1}{3^2} \right) \][/tex]
[tex]\[ E_3 = - R_H \left( \frac{1}{9} \right) \][/tex]
[tex]\[ E_3 = - 2.18 \times 10^{-18} \times \frac{1}{9} \][/tex]
[tex]\[ E_3 \approx -2.422222222222222 \times 10^{-19} \, \text{J} \][/tex]
The energy emitted by the electron as it transitions from [tex]\( n = 5 \)[/tex] to [tex]\( n = 3 \)[/tex]:
[tex]\[ \Delta E = E_3 - E_5 \][/tex]
[tex]\[ \Delta E = -2.422222222222222 \times 10^{-19} - (-8.72 \times 10^{-20}) \][/tex]
[tex]\[ \Delta E \approx -1.550222222222222 \times 10^{-19} \, \text{J} \][/tex]
The energy emitted by one mole of electrons:
[tex]\[ \text{Energy}_\text{mole} = \Delta E \times N_A \][/tex]
[tex]\[ \text{Energy}_\text{mole} = -1.550222222222222 \times 10^{-19} \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{Energy}_\text{mole} \approx -93354.38222222221 \, \text{J/mol} \][/tex]
So, the energy emitted by one mole of electrons is approximately [tex]\(-93354.38222222221 \, \text{J/mol}\)[/tex].
#### (c) Wavelength of the emitted photon
Using the relationship between the energy of a photon, its wavelength, and the constants [tex]\( h \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ \Delta E = h \nu \][/tex]
[tex]\[ \Delta E = \frac{h c}{\lambda} \][/tex]
Rearranging for the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{h c}{\Delta E} \][/tex]
Substituting the values:
[tex]\[ \lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.550222222222222 \times 10^{-19}} \][/tex]
[tex]\[ \lambda \approx 1.2830418577981652 \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ \lambda \approx 1283 \, \text{nm} \][/tex] (convert meters to nanometers by multiplying by [tex]\(10^9\)[/tex])
So, the wavelength of the emitted photon is approximately [tex]\( 1283 \, \text{nm} \)[/tex].
