Question

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is:
(A) 802nm
(B) 823nm
(C) 1882nm
(D) 1648nm

Answer 1

Hint: The equation that relates the wavelength and the transition in given atoms is
\[\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right]\]
Here, the transition occurs from ${n_1}$ to ${n_2}$ level. Value of R is constant for a given atom.

Complete step by step solution:
We know that the wavelength of the ultraviolet spectrum is smaller than the infrared region. We are given the largest wavelength of the UV spectrum of hydrogen. So, from this information, we will find the smallest wavelength of the infrared region.
-Now, for the largest wavelength in the UV spectrum corresponds to a particular transition in the hydrogen atom. We know that it corresponds to ${n_1}$=2 to ${n_2}$=1 transition. This particular transition is from the Lyman series. Now, formula that relates the wavelength, and transitions can be given as
\[\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right]\] ………………(1)
So, we can write this equation for the case in which we get wavelength 122nm as
\[\dfrac{1}{{122}} = R\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right]\]
Therefore \[\dfrac{1}{{122}} = R\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{4}} \right]\]
So, we can write that $R = \dfrac{3}{{4 \times 122nm}}$
So, we obtained the value of R which will be the same for the other transitions in hydrogen atom.
-We can say that the smallest wavelength in the infrared region corresponds to maximum energy of the series which is ${n_1} = \infty $ to ${n_2}$=3 which is involved in the Paschen series.
So, we can write the equation (1) for that as
\[\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{3^2}}} - \dfrac{1}{\infty }} \right]\] ………..(2)
 Now, we obtained the value of R earlier. So, putting its value in equation (2), we get
\[\dfrac{1}{\lambda } = \dfrac{3}{{4 \times 122}}\left[ {\dfrac{1}{9} - 0} \right]\]
So,
\[\lambda = \dfrac{{9 \times 3 \times 122}}{4} = 823.5nm\]
Thus, we can say that the closest integer for the wavelength corresponding to the smallest wavelength in the IR spectrum of hydrogen atom is 823nm.

So, the correct answer is (B).

Note: Remember that 1nm=${10^{ - 9}}$m. So, whenever the wavelength is given in any other unit of length, then carefully convert them accordingly. Note that the smallest wavelength in the IR spectrum corresponds to maximum energy of the series ${n_1} = \infty $ to ${n_2} = 3$ transition in the Paschen series.