Question

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
A) $\dfrac{7}{{29}}$
B) $\dfrac{9}{{31}}$
C) $\dfrac{5}{{27}}$
D) $\dfrac{3}{{23}}$

Answer 1

Hint: In the hydrogen-spectral series, the first lines of each series have the lowest frequency and they correspond to the longest wavelengths, as wavelength is inversely proportional to its frequency.
$
  V = f \times \lambda \\
  V = C = 3 \times {10^8} \\
  \therefore f \propto \dfrac{1}{\lambda } \\
 $

Complete step by step answer:
Velocity of light is equal to the product of wavelength and frequency. Since, velocity of light is constant, wavelength is inversely proportional to frequency.
$
\Rightarrow V = f \times \lambda \\
\Rightarrow V = C = 3 \times {10^8} \\
  \therefore f \propto \dfrac{1}{\lambda } \\
 $
Thus, the spectral lines corresponding to the longest wavelengths are in turn, the ones with the shortest frequency i.e. shortest electron jump. This is nothing, but the first line of each series since the electron has just advanced to the most minimum level of one.
Hence,
First member of Lyman Series : ${n_1} = 1 \to {n_2} = 2$
First member of Balmer Series: ${n_1} = 2 \to {n_2} = 3$
Rydberg’s equation for the transition of an electron from ${n_2} \to {n_1}$:
$\Rightarrow$ \[\upsilon = \dfrac{1}{\lambda } = R{Z^2}\left\{ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right\}\]
where
$
\Rightarrow \upsilon = wavenumber \\
\lambda = wavelength \\
$
$R$ = Rydberg’s constant, equal to $1.097 \times {10^{ - 7}}{m^{ - 1}}$
$Z$ = atomic number of the element
Applying the Rydberg’s formula for the first line of Lyman series ${n_1} = 1 \to {n_2} = 2$, we get:
\[
  {\upsilon _1} = \dfrac{1}{{{\lambda _1}}} = R{Z^2}\left\{ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right\} \\
  Substituting, \\
  {\upsilon _1} = \dfrac{1}{{{\lambda _1}}} = R{Z^2}\left\{ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right\} \\
  Solving, \\
\Rightarrow \dfrac{1}{{{\lambda _1}}} = R{Z^2}\left\{ {\dfrac{1}{1} - \dfrac{1}{4}} \right\} \\
\Rightarrow \dfrac{1}{{{\lambda _1}}} = R{Z^2}\left\{ {\dfrac{3}{4}} \right\} \\
\Rightarrow {\lambda _1} = \dfrac{{4R{Z^2}}}{3} \to (1) \\
 \]
Now, apply the Rydberg’s formula for the first line of Balmer series ${n_1} = 2 \to {n_2} = 3$, we get:
\[
  {\upsilon _2} = \dfrac{1}{{{\lambda _2}}} = R{Z^2}\left\{ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right\} \\
  Substituting, \\
  {\upsilon _2} = \dfrac{1}{{{\lambda _2}}} = R{Z^2}\left\{ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right\} \\
  Solving, \\
\Rightarrow \dfrac{1}{{{\lambda _2}}} = R{Z^2}\left\{ {\dfrac{1}{4} - \dfrac{1}{9}} \right\} \\
\Rightarrow \dfrac{1}{{{\lambda _2}}} = R{Z^2}\left\{ {\dfrac{5}{{36}}} \right\} \\
\Rightarrow {\lambda _2} = \dfrac{{36R{Z^2}}}{5} \to (2) \\
 \]
From (1) and (2), the ratio of wavelengths are:
$\Rightarrow$ \[\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{4R{Z^2}}}{3} \div \dfrac{{36R{Z^2}}}{5} = \dfrac{{\not 4\not R{{\not Z}^2}}}{3} \times \dfrac{5}{{\not 3\not 6\not R{{\not Z}^2}9}} = \dfrac{5}{{27}}\]

Hence, the correct option is Option C.

Note: Sometimes, you could be confused in the Rydberg’s formula as to whether it is \[\left\{ {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right\}or\left\{ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right\}\], so it’s better to remember by lower one comes first and higher one comes later.