Question

The masses of photons corresponding to the first lines of Brackett and Balmer series of the atomic spectrum of hydrogen are in the ratio of:
A.$\dfrac{5}{{27}}$
B.$\dfrac{1}{{320}}$
C.$\dfrac{{81}}{{500}}$
D.$\dfrac{{500}}{{81}}$

Answer 1

The energy difference between the various levels of Bohr’s model and the wavelengths of absorbed or emitted photon is given by the Rydberg formula:
$\dfrac{{\text{1}}}{{{\lambda }}}{\text{ = R}}{{\text{Z}}^{\text{2}}}\left( {\dfrac{{\text{1}}}{{{{\text{n}}_{\text{1}}}^{\text{2}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{n}}_{\text{2}}}^{\text{2}}}}} \right)$

Complete step by step answer:
The atomic spectrum of hydrogen depicts the electronic transitions of the hydrogen atom. Each electronic transition series corresponds to a particular series and corresponding energy level. Following are the spectra series of the hydrogen atom:
Lyman series is the series where the electronic transition takes place from higher energy level states such as (n=2,3,4…) to n=1 energy state.
Balmer series is the series when electronic transition takes place from higher energy level states such as (n=3,4,5…) to n=2 energy states.
The Paschen series takes place from higher energy level states such as (n=4,5…) to n=3 energy states.
Brackett series takes place from higher energy level states such as (n=5,6…) to n=4 energy states.
To find the ratio of masses of Brackett and Balmer series. To find it first we need to know that wavelength is inversely proportional to the mass. This is depicted from the DeBroglie formula:

$${{\lambda = }}\dfrac{{\text{h}}}{{{\text{mv}}}} \\\ \Rightarrow {{\lambda }} \propto \dfrac{{\text{1}}}{{\text{m}}} \\\$$

Now, we need to find the energy difference,
The energy difference between the various levels of Bohr’s model and the wavelengths of absorbed or emitted photon is given by the Rydberg formula:
$\dfrac{{\text{1}}}{{{\lambda }}}{\text{ = R}}{{\text{Z}}^{\text{2}}}\left( {\dfrac{{\text{1}}}{{{{\text{n}}_{\text{1}}}^{\text{2}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{n}}_{\text{2}}}^{\text{2}}}}} \right)$
Where ${{\lambda }}$ is the wavelength (nm),
Z is the atomic number (For hydrogen atom Z =1),
R is the Rydberg constant which is equal to ${\text{1}}{\text{.097 x 1}}{{\text{0}}^{\text{7}}}{\text{ c}}{{\text{m}}^{{\text{ - 1}}}}$
${{\text{n}}_{\text{1}}}$ is the lowest energy level, and
${{\text{n}}_{\text{2}}}$is the highest energy level.
For Brackett series, ${{\text{n}}_{\text{1}}} = 4$
$\dfrac{{\text{1}}}{{{{{\lambda }}_{\text{1}}}}}{\text{ = R}}\left( {\dfrac{{\text{1}}}{{{4^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{5^{\text{2}}}}}} \right){\text{ = }}\dfrac{{{\text{9R}}}}{{400}}$
For Balmer series, ${{\text{n}}_{\text{1}}} = 2$
$\dfrac{{\text{1}}}{{{{{\lambda }}_2}}}{\text{ = R}}\left( {\dfrac{{\text{1}}}{{{{\text{2}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{3}}^{\text{2}}}}}} \right){\text{ = }}\dfrac{{{\text{5R}}}}{{36}}$
${{{\lambda }}_{\text{1}}}{\text{:}}{{{\lambda }}_{\text{2}}}{\text{ = }}\dfrac{{{\text{5R}}}}{{{\text{36}}}}{\text{:}}\dfrac{{{\text{9R}}}}{{{\text{400}}}} = \dfrac{{500}}{{81}}$
Since masses are inversely proportional to the wavelengths.
So, the masses are in the ratio of ${{\text{m}}_{\text{1}}}{\text{:}}{{\text{m}}_{\text{2}}}{\text{ = 81:500}}$
The masses of photons corresponding to the first lines of the Brackett and Balmer series of the atomic spectrum of hydrogen are in the ratio of $\dfrac{{81}}{{500}}$.

So, the correct option is C.

Note:
The masses of photons are inversely proportional to the wavelength but directly proportional to the frequency of the photon. Brackett and Balmer’s series occur in the infra-red and visible region of the spectrum respectively.