Felix Castory
Jun 23
Electromagnetic radiation was emitted in hydrogen spectrum as the result of electron transition from fifth bohr orbit to visible part, calculate; wavelength of the emitted radiation ?
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ChatGPT
Jun 23
To calculate the wavelength of the emitted radiation, we can use the formula:
\[ \lambda = \frac{hc}{E} \]
Where:
- λ is the wavelength of the emitted radiation
- h is the Planck's constant (6.626 x 10^-34 J s)
- c is the speed of light (3.00 x 10^8 m/s)
- E is the energy difference between the initial and final energy levels of the electron transition
The energy difference between the fifth and visible part of the hydrogen spectrum can be calculated using the formula:
\[ E = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} \]
Where:
- n_f is the final energy level (in this case, 2 for the visible part)
- n_i is the initial energy level (in this case, 5 for the fifth Bohr orbit)
Substitute the values into the formula:
\[ E = -\frac{13.6}{2^2} + \frac{13.6}{5^2} \]
\[ E = -3.4 + 0.544 \]
\[ E = -2.856 \text{ eV} \]
Convert the energy to joules:
\[ E = -2.856 \times 1.6 \times 10^{-19} \]
\[ E = -4.5696 \times 10^{-19} \text{ J} \]
Now, substitute the values into the first formula to calculate the wavelength:
\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{4.5696 \times 10^{-19} \text{ J}} \]
\[ \lambda = \frac{1.9878 \times 10^{-25}}{4.5696 \times 10^{-19}} \]
\[ \lambda = 4.35 \times 10^{-7} \text{ m} \]
Therefore, the wavelength of the emitted radiation is 435 nm.
\[ \lambda = \frac{hc}{E} \]
Where:
- λ is the wavelength of the emitted radiation
- h is the Planck's constant (6.626 x 10^-34 J s)
- c is the speed of light (3.00 x 10^8 m/s)
- E is the energy difference between the initial and final energy levels of the electron transition
The energy difference between the fifth and visible part of the hydrogen spectrum can be calculated using the formula:
\[ E = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} \]
Where:
- n_f is the final energy level (in this case, 2 for the visible part)
- n_i is the initial energy level (in this case, 5 for the fifth Bohr orbit)
Substitute the values into the formula:
\[ E = -\frac{13.6}{2^2} + \frac{13.6}{5^2} \]
\[ E = -3.4 + 0.544 \]
\[ E = -2.856 \text{ eV} \]
Convert the energy to joules:
\[ E = -2.856 \times 1.6 \times 10^{-19} \]
\[ E = -4.5696 \times 10^{-19} \text{ J} \]
Now, substitute the values into the first formula to calculate the wavelength:
\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{4.5696 \times 10^{-19} \text{ J}} \]
\[ \lambda = \frac{1.9878 \times 10^{-25}}{4.5696 \times 10^{-19}} \]
\[ \lambda = 4.35 \times 10^{-7} \text{ m} \]
Therefore, the wavelength of the emitted radiation is 435 nm.
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Google Gemini AI
Jun 23
The energy of an electron in a hydrogen atom is given by:
$$E_n = -13.6/n^2 \text{ eV}$$
where $n$ is the principal quantum number.
The wavelength of the emitted radiation is given by:
$$\lambda = hc/E$$
where $h$ is Planck's constant, $c$ is the speed of light, and $E$ is the energy of the emitted radiation.
The energy of the emitted radiation is the difference in energy between the two energy levels involved in the transition:
$$E = E_5 - E_2 = -13.6/5^2 - (-13.6/2^2) = 10.2 \text{ eV}$$
Therefore, the wavelength of the emitted radiation is:
$$\lambda = hc/E = (1240 \text{ eV nm})/10.2 \text{ eV} = 121.6 \text{ nm}$$
This wavelength is in the visible part of the spectrum, and corresponds to the red Balmer line.
$$E_n = -13.6/n^2 \text{ eV}$$
where $n$ is the principal quantum number.
The wavelength of the emitted radiation is given by:
$$\lambda = hc/E$$
where $h$ is Planck's constant, $c$ is the speed of light, and $E$ is the energy of the emitted radiation.
The energy of the emitted radiation is the difference in energy between the two energy levels involved in the transition:
$$E = E_5 - E_2 = -13.6/5^2 - (-13.6/2^2) = 10.2 \text{ eV}$$
Therefore, the wavelength of the emitted radiation is:
$$\lambda = hc/E = (1240 \text{ eV nm})/10.2 \text{ eV} = 121.6 \text{ nm}$$
This wavelength is in the visible part of the spectrum, and corresponds to the red Balmer line.
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