Calculate the shortest and longest wavelength $(\lambda )$ in the Lyman series of Hydrogen atoms.
Rydberg's constant = $109,677c{m^{ - 1}}$
Answer
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Hint: The atomic structure of hydrogen consists of a large number of spectral lines which are grouped into five series which are Lyman , Balmer , Paschen ,Brackett and Pfund . In the given question we have to find the wavelength in the Lyman series .
Complete step by step answer:
When an electron jumps from higher energy levels $(n > 1)$ to n=1 energy level , the group of lines produced are termed as Lyman series .
To calculate the shortest and longest wavelength in Lyman series of Hydrogen atom we will use Rydberg formula which is given by
$\dfrac{1}{\lambda } = R(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$
where , $\lambda $ = wavelength , R = Rydberg's constant ,
${n_1}$ = the lower energy level to which the electron jumps ( in case of Lyman series ${n_1} = 1$ ) ${n_2}$ = The higher energy level from which the electron jumps .
Now , first we will calculate the shortest wavelength , For the wavelength to be minimum the energy difference in two states showing transition should be maximum , that is ${n_2} = \infty $ .
So on substituting the values in the rydberg's formula we get ,
$\dfrac{1}{\lambda } = 109677(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}})$
$ \Rightarrow \lambda = \dfrac{1}{{109677}} = 911.7{A^ \circ }$
Now to calculate the longest wavelength the energy difference in two states showing transition should be maximum , that is ${n_2} = 2$ .
So on substituting the values we get
$\dfrac{1}{\lambda } = 109677(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}})$
$ \Rightarrow \lambda = \dfrac{4}{{3 \times 109677}} = 1215.69{A^ \circ }$
Therefore , the shortest and longest wavelengths in Lyman series of hydrogen are $911.7{A^ \circ }$ and $1215.69{A^ \circ }$ respectively .
Note:
The line spectrum of hydrogen is explained by Bohr's model . According to Bohr's model , an electron neither emits or absorbs energy as long as it is in the same energy level but it emits or absorbs energy when it jumps from one level to another .
Complete step by step answer:
When an electron jumps from higher energy levels $(n > 1)$ to n=1 energy level , the group of lines produced are termed as Lyman series .
To calculate the shortest and longest wavelength in Lyman series of Hydrogen atom we will use Rydberg formula which is given by
$\dfrac{1}{\lambda } = R(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$
where , $\lambda $ = wavelength , R = Rydberg's constant ,
${n_1}$ = the lower energy level to which the electron jumps ( in case of Lyman series ${n_1} = 1$ ) ${n_2}$ = The higher energy level from which the electron jumps .
Now , first we will calculate the shortest wavelength , For the wavelength to be minimum the energy difference in two states showing transition should be maximum , that is ${n_2} = \infty $ .
So on substituting the values in the rydberg's formula we get ,
$\dfrac{1}{\lambda } = 109677(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}})$
$ \Rightarrow \lambda = \dfrac{1}{{109677}} = 911.7{A^ \circ }$
Now to calculate the longest wavelength the energy difference in two states showing transition should be maximum , that is ${n_2} = 2$ .
So on substituting the values we get
$\dfrac{1}{\lambda } = 109677(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}})$
$ \Rightarrow \lambda = \dfrac{4}{{3 \times 109677}} = 1215.69{A^ \circ }$
Therefore , the shortest and longest wavelengths in Lyman series of hydrogen are $911.7{A^ \circ }$ and $1215.69{A^ \circ }$ respectively .
Note:
The line spectrum of hydrogen is explained by Bohr's model . According to Bohr's model , an electron neither emits or absorbs energy as long as it is in the same energy level but it emits or absorbs energy when it jumps from one level to another .
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