Question

The wavelength of the radiation emitted by hydrogen when compared to $H{e^ + }$ is:
A.2 times that of $H{e^ + }$ ion
B.3 times that of $H{e^ + }$ ion
C.4 times that of $H{e^ + }$ ion
D.same as $H{e^ + }$

Answer 1

Wavelength is the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.
The wavelength by the Rydberg can be given by:
$\Rightarrow$
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where $\lambda$ is the wavelength, ${n_1}^2$ and ${n_2}^2$ be the states where ${n_1}$ is the state where the value is one as it is the stationary state and ${n_2}$is infinity as stated in the question. Further,$Z$ is the atomic number of the substance. Using this formula, we compared the wavelength of the radiation emitted by hydrogen to $H{e^ + }$.

Complete answer: Rydberg's formula for hydrogen like element
$\Rightarrow$
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where $R$ is Rydberg's constant$= 1.097 \times {10^7}{m^{ - 1}}$.
$\therefore \lambda \propto \dfrac{1}{{{Z^2}}}$---(1)
Since the atomic number of $H$is one and atomic number of $H{e^ + }$is two
Hence, Atomic number of $H{e^ + }$ is double of $H$.
Let ${\lambda _H}$ and ${\lambda _{H{e^ + }}}$be the wavelength of $H$and $H{e^ + }$respectively.
Then from the equation (1), we get
$\dfrac{{{\lambda _H}}}{{{\lambda _{H{e^ + }}}}} \propto \dfrac{{\left( {\dfrac{1}{{{1^2}}}} \right)}}{{\left( {\dfrac{1}{{{2^2}}}} \right)}} = 4$
\Rightarrow $$$${\lambda _H} = 4{\lambda _{H{e^ + }}}
Hence, the correct option is C.

Note:
Note that the wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to the stationary state. Also note that after the Balmer series, Rydberg got the improvised method of finding the wavelength in the hydrogen emission spectrum. We say that the formula given by Rydberg directly generalises the equations used to calculate the wavelength of the hydrogen spectral lines.