Question

what is wavelength of spectral line in balmer series where transition occur from 5th orbit​

Answer 1

Answer:

plz mark brainlist

Explanation:

Whenever an electron in a hydrogen atom jumps from higher energy level to the lower energy level to the lower energy level, the difference in energies of the two levels is emitted as a radiation of particular wavelength. It is called a spectral line. As the wavelength of the spectral line depends upon the two orbits (energy levels) between which the transition of electron takes place, various spectral lines are obtained. The different wavelengths constitute spectral series which are the characteristic of the atoms emitting them. The following are the spectral series of hydrogen atom :

(i) Lyman series : When the electron jumps from any of the outer orbits to the first orbit, the spectral lines emitted are in the ultraviolet region of the spectrum and they are said to form a series called Lyman series

Answer 2

Given:

$n=5$

To Find: Wavelength of the spectral line in Balmer series where the transition occurs from 5th orbit​

Solution:

The formula of the wavelength of a photon in the Balmer series is given as:

$\[\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} \right)\]$

As it is given that the transition occurs from the 5th orbit​, therefore, $n=5$.

Thus,

$\[\begin{gathered} \frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{5^2}}}} \right) \hfill \\ \frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{{25}}} \right) \hfill \\ \frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{{25}}} \right) \hfill \\ \frac{1}{\lambda } = R \times \frac{{21}}{{100}} \hfill \\ \end{gathered} \]$

Therefore, the wavelength can be given as:

$\[\begin{gathered} \lambda = \frac{{100}}{{21R}} \hfill \\ \lambda = \frac{{100}}{{21 \times 1.097 \times {{10}^7}}} \hfill \\ \end{gathered} \]$

Therefore,

$\[\lambda = 4.34/{10^7} = 434\,nm\]$

Hence, the wavelength of the spectral line in the Balmer series where the transition occurs from the 5th orbit​ is $\[434\,nm\]$.

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