Question

In hydrogen atom spectrum, frequency of $ 2.7\times 10^{15}Hz $ of electromagnetic wave is emitted when transmission takes place from to $1$. If it moves from $3$ to $1$, the frequency emitted will be

• A. $ 3.2\times 10^{15}Hz $
• B. $ 32\times 10^{15}Hz $
• C. $ 1.6\times 10^{15}Hz $
• D. $ 16\times 10^{15}Hz $

Answer 1

Answer: (A)

The frequency $v$ of the emitted electromagnetic radiation, when a hydrogen atom de-excites from level $n_{2}$ to $n_{1}$ is
$v=R c Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
where $n_{1}$ is lower level, $n_{2}$ is higher level, $R$ is Rydberg constant, $c$ is velocity of light and $Z$ is atomic number of atom.
When transition takes place from
$. n_{2}=2$ to $n_{1}=1$, then
$2.7 \times 10^{15}=R c Z^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ ... (i)
When transition takes places from $n_{2}=3$ to $n_{1}=1$,
let frequency be $v$.
$\therefore v=R c Z^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right) ...$(ii)
From Eqs. (i) and (ii), we get
$v =\frac{32 \times 2.7 \times 10^{15}}{27} $
$=3.2 \times 10^{15} Hz$