Question

Calculate the shortest and longest wavelengths of the Balmer series of Hydrogen atoms. Given $R = 1.097 \times {10^7} /m.$

Answer 1

In this question use the direct formula for the Balmer series of hydrogen spectrum that is

According to Balmer series of hydrogen $(H_2)$ spectrum the wavelength is given as, $\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ where R is molar gas constant, $n_1$ and $n_2$ are the spectrum numbers where, $n_1$ is always less than $n_2$. Consider the fact that for shortest wavelength ${n_2}$ should be infinity and for longest wavelength values of $n_1$ and $n_2$ are 1 and 2. This will help solve this problem.

Formula used:

$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$............................ (1)

Where, $\lambda$ = wavelength, R = molar gas constant $= 1.097 \times {10^7}/m$ and $n_1$ and $n_2$ are the spectrum numbers where, $n_1$ is always less than $n_2$.

Where $n_1$ varies from 1 to infinity and $n_2$ varies from 2 to infinity.

Complete step-by-step solution:

For shorter wavelength: For shorter wavelengths the value of $n_1$ should be least and the value of $n_2$ should be maximum. The values of $n_1$ and $n_2$ are 2 and infinity ($\infty$), as $\left[ {\dfrac{1}{\infty } = 0} \right]$

Now substitute all the values in equation (1) we have,

$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right]$

Now simplify this we have,

$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}\left[ {\dfrac{1}{4} - 0} \right] = \dfrac{1.097 \times {10^7}}{4}$

$\Rightarrow \lambda = \dfrac{4}{{1.097 \times {{10}^7}}} = 3.646 \times {10^{ - 7}}$ meter.

$\Rightarrow \lambda = 3646 \times {10^{ - 10}}$ meter.

Now as we all know that $1{A^o} = 1 \times {10^{ - 10}}$m

Therefore, $\lambda = 3646{A^o}$

$\therefore$ The shortest wavelength is $3646{A^o}$.

For longest wavelength: For the longest wavelength the value of $n_1$ should be least and the value of $n_2$ also should be least.The values of $n_1$ and $n_2$ are 2 and 3.Now substitute all the values in equation (1) we have,

$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right]$

Now simplify this we have,

$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}\left[ {\dfrac{1}{4} - \dfrac{1}{9}} \right] = 1.097 \times {10^7}\left( {\dfrac{5}{36}} \right)$

$\Rightarrow \lambda = \dfrac{{\dfrac{36}{5}}}{{1.097 \times {{10}^7}}} = \left( {\dfrac{36}{5}} \right)0.91157 \times {10^{ - 7}} = 6.563 \times {10^{ - 7}}$ meter.

$\Rightarrow \lambda = 6563 \times {10^{ - 10}}$ meter.

Now as we all know that $1{A^o} = 1 \times {10^{ - 10}}$m

Therefore, $\lambda = 6563{A^o}$

$\therefore$ The longest wavelength is $6563{A^o}$.

Note: Talking about a series of spectral emission lines for the hydrogen atom that can be achieved from the result of electron transmission from some higher level even down to the energy level of principal quantum 2 can be termed as a Balmer series.