Question

The first three spectral lines of H-atom in the Balmer series are given λ1, λ2, λ3 considering the Bohr atomic model, the wavelengths of first and third spectral lines $\left( \frac{{{\lambda }_{1}}}{{{\lambda }_{3}}} \right)$ are related by a factor of approximately $'x'\times {{10}^{-1}}$. The value of x, to the nearest integer, is _______?

Answer 1

Hint: From the Bohr atomic model the frequency can be found out and from which the formula of wave number can be easily calculated. Using that equation of wave number and substituting the values of higher and lower excited states of the Balmer series we are able to solve the question. Remember wave number is the number of waves passing per unit length. It is just the inverse of wavelength.

Formula used:
Equation of wavenumber,\[\frac{1}{\lambda }=R{{z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where $n_1$ is the ground state and $n_2$ is the higher excited state.

Complete answer:
Using equation for wavenumber,\[\frac{1}{\lambda }=R{{z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where $n_1$ is ground state and $n_2$ is higher excited state. For Balmer series $n_1$= 2 and $n_2$=3 for the first line of the Balmer series. For the second line of the Balmer series, $n_1$=2 and $n_2$ =4 and for the third line of Balmer series $n_1$=2 and $n_2$=5. For hydrogen atomic number, z=1.

Therefore, for first line of Balmer series on putting values in the equation
\[\] $\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{4}-\frac{1}{9} \right)$

The wavelength of first line of Balmer series is:
$\frac{1}{{{\lambda }_{1}}}=\frac{5R}{36}$

For third line of Balmer series
$\frac{1}{{{\lambda }_{3}}}=R\left( \frac{1}{4}-\frac{1}{25} \right)$

The wavelength of third line of Balmer series is
$\dfrac {1}{\lambda_3}=\dfrac {21~R}{100}$

Now according to question,
\[\therefore \frac{{{\lambda }_{1}}}{{{\lambda }_{3}}}=\frac{5R}{36}\times \frac{100}{21R}\]
\[\therefore \frac{{{\lambda }_{1}}}{{{\lambda }_{3}}}=15.12\times {{10}^{-1}}\]
We wanted the value of x, so the answer is x=15

Note:Here for hydrogen z=1 so it can be neglected. Hydrogen spectrum includes five series. Lyman series, Balmer series, Paschen series, Brackett series, Pfund series. The Lyman series is obtained in the ultraviolet region and the Balmer series is obtained in the visible region. As the Balmer series is obtained in the visible region it is visible to naked eyes. Paschen, Brackett and Pfund series are obtained in the infrared region. The transition occurs between n=2 to n=3, 4, 5,...∞. R is called Rydberg constant. The value of Rydberg constant is $1.097\times {{10}^{7}}{{m}^{-1}}$.