Solveeit Logo
Login

Question

Question: What is the shortest wavelength present in the Paschen series of spectral lines....

What is the shortest wavelength present in the Paschen series of spectral lines.

Explanation

Solution

The Paschen series are the series of lines in the spectrum which are emitted when the atoms of the hydrogen correspond to transitions between the different states with principal quantum number n=3 and higher states.

Formula used: The formula of the Rydberg’s formula is given by,
⇒hcλ=21⋅76×10−19×(1n12−1n22)\Rightarrow \dfrac{{hc}}{\lambda } = 21 \cdot 76 \times {10^{ - 19}} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)

Where the Planck’s constant is h, the speed of light is c, the wavelength is λ\lambda and the number of the spectral lines is n1{n_1}andn2{n_2}.

Complete answer:
It is asked in the problem what is the shortest wavelength present in the Paschen series of spectral lines.
The Paschen series is the spectral line which corresponds to the electrons which transits from the higher energy line to the lower energy line i.e. n=3n = 3.

The formula of the Rydberg’s formula is given by,

⇒hcλ=21⋅76×10−19×(1n12−1n22) \Rightarrow \dfrac{{hc}}{\lambda } = 21 \cdot 76 \times {10^{ - 19}} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)

Where the Planck’s constant is h, the speed of light is c, the wavelength is λ\lambda and the number of the spectral lines is n1{n_1}andn2{n_2}.

Here the spectral series will start from n1=3{n_1} = 3 and as we need to calculate the value of least wavelength of spectrum therefore, n2=∞{n_2} = \infty replacing these values in the Rydberg’s formula we get.

⇒hcλ=21⋅76×10−19×(1n12−1n22) \Rightarrow \dfrac{{hc}}{\lambda } = 21 \cdot 76 \times {10^{ - 19}} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)
⇒hcλ=21⋅76×10−19×(132−1∞2)\Rightarrow \dfrac{{hc}}{\lambda } = 21 \cdot 76 \times {10^{ - 19}} \times \left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)
⇒hcλ=21⋅76×10−19×(132−0)\Rightarrow \dfrac{{hc}}{\lambda } = 21 \cdot 76 \times {10^{ - 19}} \times \left( {\dfrac{1}{{{3^2}}} - 0} \right)
⇒hcλ=21⋅76×10−19×19\Rightarrow \dfrac{{hc}}{\lambda } = 21 \cdot 76 \times {10^{ - 19}} \times \dfrac{1}{9}
⇒hcλ=2⋅418×10−19\Rightarrow \dfrac{{hc}}{\lambda } = 2 \cdot 418 \times {10^{ - 19}}
⇒λ=hc2⋅418×10−19\Rightarrow \lambda = \dfrac{{hc}}{{2 \cdot 418 \times {{10}^{ - 19}}}}

Replacing the values of the Planck’s constant and speed of light we get,

⇒λ=hc2⋅418×10−19 \Rightarrow \lambda = \dfrac{{hc}}{{2 \cdot 418 \times {{10}^{ - 19}}}}
⇒λ=6⋅626×10−34×3×1082⋅418×10−19\Rightarrow \lambda = \dfrac{{6 \cdot 626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2 \cdot 418 \times {{10}^{ - 19}}}}
⇒λ=6⋅626×10−34×3×108×10+192⋅418\Rightarrow \lambda = \dfrac{{6 \cdot 626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times {{10}^{ + 19}}}}{{2 \cdot 418}}
⇒λ=8⋅20×10−7m\Rightarrow \lambda = 8 \cdot 20 \times {10^{ - 7}}m
⇒λ=820×10−9m\Rightarrow \lambda = 820 \times {10^{ - 9}}m
⇒λ=820nm\Rightarrow \lambda = 820nm.

The shortest wavelength of the Paschen series is equal to λ=820nm\lambda = 820nm.

Note: When the electrons transits in different states then a spectrum of lines are recorded and these are known as Paschen series if the principal quantum number is more than n=3. The spectral series is the wavelength of the electromagnetic radiation arranged in the sequence.