Question

In hydrogen atom spectrum, frequency of $2.7\times {10}^{15}Hz$ of EM wave is emitted when transmission takes place from $2$ 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

The correct option is A $3.2\times {10}^{15}Hz$

The frequency $v$ of the emitted electromagnetic radiation, when a hydrogen atom de-excites from the level $n_2$ to $n_1$ is

$v=RC{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

When transition takes place from $n_2=2$ to $n_1=1$, then

$2.7\times {10}^{15}=RC{Z}^2(\frac{1}{1^2}-\frac{1}{2^2})$

When transition takes place from $n_2=3$ to $n_1=1$, then let the frequency be $v$.

$\therefore v=RC{Z}^2(\frac{1}{1^2}-\frac{1}{3^2})$

From equations (i) and (ii), we get

$v=\frac{32\times 2.7\times {10}^{15}}{27}=3.2\times {10}^{15}Hz$