
Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.
Answer
511.2k+ views
Hint: For the longest wavelength the change in energy between the states of electrons has to be the least.
Complete answer:
First let’s see the formula for the calculation of wavelength when an electron jumps from a specific state to another state: -
$\dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}]$
Now this formula states that the reciprocal of wavelength given by $\lambda $is directly proportional to the difference of reciprocal squares of the transitional states ${n_1} \to {n_2}$.
Now for the wavelength to be the smallest the difference $\dfrac{1}{{{n_2}^2}}$ has to be the least that is ${n_2}$has to be the largest. For ${n_2}$to be the largest it has to be infinity. The value of wavelength after this would be the reciprocal of $R{Z^2}$. The energy dispersed in the process is the highest.
Now for the wavelength to be the longest of the absorption spectrum then
First the absorption spectrum is the one in which an electron gets bombarded by a photon and then the electron jumps a few states depending on the energy of the photon particle.
So, for the absorption spectrum to be the longest, the difference $\dfrac{1}{{{n_2}^2}}$ has to be the greatest. The highest value that ${n_2}$ can assume is the second stage, which is the next numeric state after the ground state. Here the energy absorbed is the lowest.
Calculating the $\lambda $
$
\dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}] \\
\Rightarrow \dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}] \\
\Rightarrow \dfrac{1}{\lambda } = R{Z^2} \times \dfrac{3}{4} \\
\Rightarrow \lambda = 121.6nm \\
$
The longest wavelength of absorption line for hydrogen gas containing atoms in ground state is $121.6nm$.
Note:
The energy is directly proportional to the frequency and inversely proportional to the wavelength therefore the change in energy is inversely proportional to the wavelength of the absorption spectrum.
Complete answer:
First let’s see the formula for the calculation of wavelength when an electron jumps from a specific state to another state: -
$\dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}]$
Now this formula states that the reciprocal of wavelength given by $\lambda $is directly proportional to the difference of reciprocal squares of the transitional states ${n_1} \to {n_2}$.
Now for the wavelength to be the smallest the difference $\dfrac{1}{{{n_2}^2}}$ has to be the least that is ${n_2}$has to be the largest. For ${n_2}$to be the largest it has to be infinity. The value of wavelength after this would be the reciprocal of $R{Z^2}$. The energy dispersed in the process is the highest.
Now for the wavelength to be the longest of the absorption spectrum then
First the absorption spectrum is the one in which an electron gets bombarded by a photon and then the electron jumps a few states depending on the energy of the photon particle.
So, for the absorption spectrum to be the longest, the difference $\dfrac{1}{{{n_2}^2}}$ has to be the greatest. The highest value that ${n_2}$ can assume is the second stage, which is the next numeric state after the ground state. Here the energy absorbed is the lowest.
Calculating the $\lambda $
$
\dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}] \\
\Rightarrow \dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}] \\
\Rightarrow \dfrac{1}{\lambda } = R{Z^2} \times \dfrac{3}{4} \\
\Rightarrow \lambda = 121.6nm \\
$
The longest wavelength of absorption line for hydrogen gas containing atoms in ground state is $121.6nm$.
Note:
The energy is directly proportional to the frequency and inversely proportional to the wavelength therefore the change in energy is inversely proportional to the wavelength of the absorption spectrum.
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