Electrons of energies 10.2 eV and 12.09 eV can cause radiation to be emitted from hydrogen atoms. Calculate in each case: (a) the principal quantum number of the electron in the hy... Electrons of energies 10.2 eV and 12.09 eV can cause radiation to be emitted from hydrogen atoms. Calculate in each case: (a) the principal quantum number of the electron in the hydrogen atom if it is raised and (b) the wavelengths of radiation emitted if the electron drops back to the ground level.
Understand the Problem
The question is asking for calculations related to electrons with specific energies causing radiation in hydrogen atoms. Specifically, it requires us to find the principal quantum number in two scenarios and the emitted wavelengths.
Answer
(a) $n = 2$ for 10.2 eV; $n = 1$ for 12.09 eV. (b) Wavelengths: $122 \, \text{nm}$ for 10.2 eV; $103 \, \text{nm}$ for 12.09 eV.
Answer for screen readers
(a) For 10.2 eV, the principal quantum number is $n \approx 1.25$ (round to $n = 2$).
For 12.09 eV, the principal quantum number is $n \approx 1.067$ (round to $n = 1$).
(b) The wavelengths are approximately:
For 10.2 eV: $\lambda \approx 1.22 \times 10^{-7} , \text{m}$ (or 122 nm).
For 12.09 eV: $\lambda \approx 1.03 \times 10^{-7} , \text{m}$ (or 103 nm).
Steps to Solve
- Calculate the Principal Quantum Number for 10.2 eV
The energy difference between the levels in a hydrogen atom is given by the equation:
$$ E_n = - \frac{13.6 , \text{eV}}{n^2} $$
To find the principal quantum number $n$ for the energy of 10.2 eV, we set up the equation:
$$ 10.2 = \frac{13.6}{n^2} $$
Rearranging gives:
$$ n^2 = \frac{13.6}{10.2} $$
Now, calculate $n$.
- Calculate the Principal Quantum Number for 12.09 eV
Using the same formula as above for 12.09 eV:
$$ 12.09 = \frac{13.6}{n^2} $$
Rearranging gives:
$$ n^2 = \frac{13.6}{12.09} $$
Now, calculate $n$.
- Calculate Wavelength for 10.2 eV
The wavelength emitted can be calculated using the energy-wavelength relationship:
$$ E = \frac{hc}{\lambda} $$
where $h$ is Planck's constant ($6.626 \times 10^{-34} , \text{Js}$) and $c$ is the speed of light ($3.00 \times 10^8 , \text{m/s}$). Rearranging gives:
$$ \lambda = \frac{hc}{E} $$
Insert $E = 10.2 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV}$ to find the wavelength.
- Calculate Wavelength for 12.09 eV
Using the same formula for 12.09 eV:
$$ \lambda = \frac{hc}{E} $$
Insert $E = 12.09 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV}$ to find the wavelength.
(a) For 10.2 eV, the principal quantum number is $n \approx 1.25$ (round to $n = 2$).
For 12.09 eV, the principal quantum number is $n \approx 1.067$ (round to $n = 1$).
(b) The wavelengths are approximately:
For 10.2 eV: $\lambda \approx 1.22 \times 10^{-7} , \text{m}$ (or 122 nm).
For 12.09 eV: $\lambda \approx 1.03 \times 10^{-7} , \text{m}$ (or 103 nm).
More Information
The principal quantum number corresponds to the energy level of the electron in the atom. Higher energy levels (larger $n$ values) correspond to higher energy electrons. The emitted wavelengths can be found using the energy-wavelength relationship, demonstrating the connection between energy levels and the electromagnetic spectrum.
Tips
- Forgetting to convert eV to Joules when calculating the wavelength. Always use the conversion factor ($1 , \text{eV} = 1.602 \times 10^{-19} , \text{J}$).
- Rounding errors in calculating $n$. Always take care to use appropriate significant figures.
