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
$v=Rc{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
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
$2.7\times 10^{15}=Rc{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ ... (i)
When transition takes places from $n_2=3$ to $n_1=1$,
let frequency be $v$.
$\therefore v=Rc{Z}^2\left(\frac{1}{1^2}-\frac{1}{3^2}\right)...$(ii)
From Eqs. (i) and (ii), we get
$v=\frac{32\times 2.7\times 10^{15}}{27}$
$=3.2\times 10^{15}Hz$