
The short wavelength limit for the Lyman series of the hydrogen spectrum is $913.4\,\mathop A\limits^ \circ $. Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum.
Answer
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Hint: The Rydberg’s formula for the wavelength of spectral lines in the hydrogen spectrum can be used. Considering the upper value of the principal quantum number as for infinite then there is no value for it. By finding the value of Rydberg’s constant and explaining it for the Balmer series, then we can get the short wavelength limit for the Blamer series of the Hydrogen spectrum.
Complete step by step solution:
The Rydberg equation for hydrogen is given as,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where,
$R$ is the Rydberg constant
$n$ is the principal quantum number
So, the short wavelength limit ${\lambda _L}$ for the given Lyman series be,
$\dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{{\left( {{1^2}} \right)}} - \dfrac{1}{{\left( {{\infty ^2}} \right)}}} \right) = R$
By solving the above equation, we can get
$R = \dfrac{1}{{913.4}}{A^{ - 1}}$
So, the value of $R$ is known.
Now, the shorter wavelength limit for Balmer series be,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{\left( {{2^2}} \right)}} - \dfrac{1}{{\left( {{\infty ^2}} \right)}}} \right) = \dfrac{R}{2}$
Thus,
${\lambda _B}$ be the shorter wavelength limit of the Balmer series.
So, the shorter wavelength limit for Balmer series be,
${\lambda _B} = \dfrac{{{L_1}}}{R}$
Substituting the value of $R$ in the above equation,
${\lambda _B} = 4 \times 913.4\,\mathop A\limits^ \circ $
By multiplying the above values,
${\lambda _B} = 3653.6\,\mathop A\limits^ \circ $
$\therefore$ The shorter wavelength limit for the Balmer series of the hydrogen spectrum be $3653.6\,\mathop A\limits^ \circ $.
Additional information:
The Lyman series is a hydrogen spectral series of transition and results in the ultraviolet emission of lines of hydrogen atoms as the electron descends from $n \geqslant 2$ to $n = 1$.
The Balmer series of the spectral lines of the hydrogen atom is the result of the transition of an electron from a higher level down to the energy level with the principal quantum number $2$.
Note:
The Balmer series is commonly useful in astronomy because these Balmer lines appear in numerous stellar objects because of the abundance of hydrogen in the universe. By finding the value of Rydberg’s constant and explaining it for the Balmer series, then we can get the short wavelength limit for the Blamer series of the Hydrogen spectrum.
Complete step by step solution:
The Rydberg equation for hydrogen is given as,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where,
$R$ is the Rydberg constant
$n$ is the principal quantum number
So, the short wavelength limit ${\lambda _L}$ for the given Lyman series be,
$\dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{{\left( {{1^2}} \right)}} - \dfrac{1}{{\left( {{\infty ^2}} \right)}}} \right) = R$
By solving the above equation, we can get
$R = \dfrac{1}{{913.4}}{A^{ - 1}}$
So, the value of $R$ is known.
Now, the shorter wavelength limit for Balmer series be,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{\left( {{2^2}} \right)}} - \dfrac{1}{{\left( {{\infty ^2}} \right)}}} \right) = \dfrac{R}{2}$
Thus,
${\lambda _B}$ be the shorter wavelength limit of the Balmer series.
So, the shorter wavelength limit for Balmer series be,
${\lambda _B} = \dfrac{{{L_1}}}{R}$
Substituting the value of $R$ in the above equation,
${\lambda _B} = 4 \times 913.4\,\mathop A\limits^ \circ $
By multiplying the above values,
${\lambda _B} = 3653.6\,\mathop A\limits^ \circ $
$\therefore$ The shorter wavelength limit for the Balmer series of the hydrogen spectrum be $3653.6\,\mathop A\limits^ \circ $.
Additional information:
The Lyman series is a hydrogen spectral series of transition and results in the ultraviolet emission of lines of hydrogen atoms as the electron descends from $n \geqslant 2$ to $n = 1$.
The Balmer series of the spectral lines of the hydrogen atom is the result of the transition of an electron from a higher level down to the energy level with the principal quantum number $2$.
Note:
The Balmer series is commonly useful in astronomy because these Balmer lines appear in numerous stellar objects because of the abundance of hydrogen in the universe. By finding the value of Rydberg’s constant and explaining it for the Balmer series, then we can get the short wavelength limit for the Blamer series of the Hydrogen spectrum.
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