Question

If the largest wavelength of a certain series of hydrogen spectrum is \[1875nm\]. Then smallest wavelength of that series will be:
A. \[1430nm\]
B. \[820nm\]
C. \[656nm\]
D. \[1150nm\]

Answer 1

Hint: Hydrogen spectrum has numerous series. Each series has its own wavelength. If the energy required for the specific transition is small, then that transition shows the largest wavelength. If the energy required for a specific transition is more, then that transition shows the smallest wavelength.

Formula used:
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}}} \right)$
Where,
$\lambda $=wavelength
$R$=Reynolds number which equals to $1.09 \times {10^7}{m^{ - 1}}$
${n_f}$=final level the electron reaches
${n_i}$=initial level from where the electron jumps.

Complete step by step solution:
For the case of the largest wavelength, ${n_i} = n$ and ${n_f} = n - 1$. Here the largest wavelength is $\lambda = 1875nm$. Apply all these values in the given formula,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}}} \right)$
$\dfrac{1}{{1875}} = R\left( {\dfrac{1}{{{{\left( {n - 1} \right)}^2}}} - \dfrac{1}{{{n^2}}}} \right).....................\left( 1 \right)$
For smaller wavelengths, ${n_f} = x$,${n_i} = n$and $\lambda = {\lambda _{\min }}$. Apply these values in the formula
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}}} \right)$
$\dfrac{1}{{{\lambda _{\min }}}} = R\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{n^2}}}} \right).................\left( 2 \right)$
Considering $n = \infty $, Then the equation (2) becomes,
$\dfrac{1}{{{\lambda _{\min }}}} = R\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{\infty }} \right)$
We know that $\dfrac{1}{\infty } = 0$, Then the above equation becomes,
$\dfrac{1}{{{\lambda _{\min }}}} = R\left( {\dfrac{1}{{{x^2}}}} \right)$
Substituting the values $x = 2,3...$
When $x = 2$
$\dfrac{1}{{{\lambda _{\min }}}} = 1.09 \times {10^7}\left( {\dfrac{1}{4}} \right)$
$ \Rightarrow {\lambda _{\min }} = 366.9nm$
When $x = 3$
$\dfrac{1}{{{\lambda _{\min }}}} = 1.09 \times {10^7}\left( {\dfrac{1}{9}} \right)$
$ \Rightarrow {\lambda _{\min }} = 820.58nm$
When $x = 4,{\lambda _{\min }} = \Rightarrow {\lambda _{\min }} = 1467nm$
The ${\lambda _{\min }}$ shows that if increase $x$, wavelength also increases. Hence, among the values we find, we have $820nm$ as an option.
Hence the correct option is option B.

Additional information:
(i) At room temperature, all hydrogen atoms are in a ground state. At higher temperatures or during an electric discharge, electrons can be in an excited state. Since their lifetime in the excited state is smaller, they make transitions to lower energy states and emit electromagnetic radiations of definite frequency. This means there are no restrictions on the change in the value of $n$ during a transition.
(ii) If the electron in the upper state jumps to a lower state, the energy emits during the transition is,
$h\nu = {E_i} - {E_f}$
(iii) The different spectral lines can be observed in the hydrogen spectrum when the values of states are changed. Those series are called Lyman series, Balmer series, bracket series, and Pfund series. Every series lies in different spectral regions.
(iv) Lyman series occurs in the ultraviolet region. Balmer series lies in the near-ultraviolet and the visible region, Paschen, bracket, and Pfund series lies in the infrared region.
(v) The spectra of atoms are really very complicated. But the hydrogen spectrum is very simple that’s why we are keeping it as an example.

Note:
Each series has many different wavelengths. As we know that, if the energy required for the transition is more, then the wavelength is short. So, the smallest wavelength occurs when the electron from the $n^{th}$ energy level reaches any lower level. As we don’t know the number of the nth level, so we assume it as infinity.