Name: Bohr Atom Model: Frequency of Emitted Radiation
Uploaded: 2019-07-19
Duration: 3 min 54 s
Description: Electrons jump in different energy levels. Let's understand how we can calculate the frequency of emitted and absorbed radiation as electrons transition into...

Question

Calculate the longest and shortest wavelength in the Lyman series of hydrogen emission spectrum. ( Rydberg constant

R = 1.1 \times 10^{7} m^{- 1}

)

Accepted Answer

The formula to determine the wavelength of transitions in hydrogen spectrum is as follows:

$\frac{1}{λ} = R_{H} [\frac{1}{{n^{2}}_{l o w e r}} - \frac{1}{{n^{2}}_{h i g h e r}}]$

λ is the wavelength.

RH is the Rydberg constant.

The value of Rydberg constant is 109678 $c m^{- 1}$ .

n-lower is the principle quantum of the energy level in which the electron comes.

n-higher is the principle quantum of the energy level from which the electron comes.

Lyman series arise when electrons from higher energy levels come to the energy level of principle quantum number one. So, the n-higher is n1.
So, the wavelength formula for Lyman series is as follows:

$\frac{1}{λ} = R_{H} [\frac{1}{{n^{2}}_{l o w e r}} - \frac{1}{{n^{2}}_{h i g h e r}}]$

Shortest wavelengths in the hydrogen spectrum of Lyman series means the electron is coming from the highest energy level. The highest energy level is $n_{\infty}$ .
So, the shortest wavelengths are,
Substitute 109678 cm^−1 for Rydberg constant, n1=1 and

$n_{\infty}$ For n-higher

$\Rightarrow 1 / λ = R_{H} [1 / {n^{2}}_{1} - 1 / {n_{\infty}}^{2}] \Rightarrow 1 λ = 109678 c m^{- 1} \Rightarrow λ = 1 / 109678 c m^{- 1} \Rightarrow λ = 9.11 \times 10^{- 6} c m$

So, the shortest wavelength is $9.11 \times 10^{- 6} c m$

Longest wavelengths in the hydrogen spectrum of Lyman series means the electron is coming from the next higher energy level. The next higher energy level is second so, the $n_{h i g h e r}$ Is 2

So, the longest wavelengths are,
Substitute $c m^{- 1}$ For Rydberg constant and $n_{2} f o r n_{h i g h e r}$

$\Rightarrow 1 / λ = 109678 c m^{- 1} [\frac{1}{{n^{2}}_{1}} - \frac{1}{{n_{2}}^{2}}] 1 / λ = 109678 c m^{- 1} [\frac{1}{1^{2}} - \frac{1}{2^{2}}] 1 / λ = 109678 c m^{- 1} \times \frac{3}{4} \Rightarrow 1 / λ = 82258.5 c m^{- 1} \Rightarrow λ = 1 / 82258.5 c m^{- 1} \Rightarrow λ = 1.22 \times 10^{5} c m$

So, the longest wavelength is $S o, t h e l o n g e s t w a v e l e n g t h i s$ $1.22 \times 10^{5} c m$

Calculate the longest and shortest wavelength in the Lyman series of hydrogen emission spectrum. ( Rydberg constant $R = 1.1 \times 10^{7} m^{- 1}$ )

Important Questions on Atoms

Learn and Prepare for any exam you want !

Calculate the longest and shortest wavelength in the Lyman series of hydrogen emission spectrum. ( Rydberg constant R=1.1×107 m-1)

Important Questions on Atoms

Learn and Prepare for any exam you want !

Calculate the longest and shortest wavelength in the Lyman series of hydrogen emission spectrum. ( Rydberg constant $R = 1.1 \times 10^{7} m^{- 1}$ )