
How many different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to state with principal quantum number \[n\] ?
Answer
516.9k+ views
Hint: When hydrogen atom is energetically excited then the electromagnetic radiations came out of it in a number of different ranges of their difference in wavelength, this collection of electromagnetic radiations in s spectrum is a evidence to show the quantization electronic nature of atoms, and these spectrum are known as Hydrogen spectrum.
Complete step by step answer:
Let us first understand the mechanism of the hydrogen spectrum in brief. When a hydrogen atom is energized enough then electrons get sufficient energy to jump from higher principal quantum state to lower quantum state and hence they release a significant amount of electromagnetic radiation of different wavelengths.
These wavelength are generally measured by a famous Rydberg formula which is mathematically represented as:
$\dfrac{1}{\lambda } = {R_H}{Z^2}[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}]$
Where, \[\lambda \] is the wavelength of the emitted electromagnetic radiation.
$Z$ Is the atomic number of atoms, for hydrogen atoms it is equal to one.
${R_H} = 1.0967{m^{ - 1}}$ It’s known as Rydberg constant.
${n_1}$ Is the lower principal state while ${n_2}$ is the higher principal state.
Now, Total number of transitions which can occur if atoms are excited with a principal quantum number \[n\] can be calculated as:
$N = (n - 1) + (n - 2) + .... + 3 + 2 + 1$
$\Rightarrow N = \dfrac{{(n - 1)}}{2}(n)$
$\therefore N = \dfrac{{n(n - 1)}}{2}$
Hence, the total number of different wavelengths can occur if an atom is excited with a principal quantum number \[n\] will be $N = \dfrac{{n(n - 1)}}{2}$.
Note: It should be remembered that the highest quantum state from which the electron can jump to its lower quantum state is taken as \[n\] . And from this higher state electrons can jump into lower states starting from $(n - 1)$ up to $1$ and these total number of ways are the only total possibility of different wavelengths which may occur during the excitation of hydrogen atoms.
Complete step by step answer:
Let us first understand the mechanism of the hydrogen spectrum in brief. When a hydrogen atom is energized enough then electrons get sufficient energy to jump from higher principal quantum state to lower quantum state and hence they release a significant amount of electromagnetic radiation of different wavelengths.
These wavelength are generally measured by a famous Rydberg formula which is mathematically represented as:
$\dfrac{1}{\lambda } = {R_H}{Z^2}[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}]$
Where, \[\lambda \] is the wavelength of the emitted electromagnetic radiation.
$Z$ Is the atomic number of atoms, for hydrogen atoms it is equal to one.
${R_H} = 1.0967{m^{ - 1}}$ It’s known as Rydberg constant.
${n_1}$ Is the lower principal state while ${n_2}$ is the higher principal state.
Now, Total number of transitions which can occur if atoms are excited with a principal quantum number \[n\] can be calculated as:
$N = (n - 1) + (n - 2) + .... + 3 + 2 + 1$
$\Rightarrow N = \dfrac{{(n - 1)}}{2}(n)$
$\therefore N = \dfrac{{n(n - 1)}}{2}$
Hence, the total number of different wavelengths can occur if an atom is excited with a principal quantum number \[n\] will be $N = \dfrac{{n(n - 1)}}{2}$.
Note: It should be remembered that the highest quantum state from which the electron can jump to its lower quantum state is taken as \[n\] . And from this higher state electrons can jump into lower states starting from $(n - 1)$ up to $1$ and these total number of ways are the only total possibility of different wavelengths which may occur during the excitation of hydrogen atoms.
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