Modern Physics or Chemistry?

Chemistry Level 2

Light of wavelength $12818 \text{ A}^{\circ }$ is emitted when an electron of a unielectronic species drops from $5^{\text{th}}$ to $3^{\text{rd}}$ energy level. Calculate the wavelength of the photon emitted when electron falls from $3^{\text{rd}}$ energy level to ground level.

Answer in $\text{nm}$ .

The answer is 102.5.

1 solution

Gautam Sharma
Jul 24, 2015

This is a very generalized solution. We know $E \propto \frac{1}{n^2}$ where $n$ is the orbit in which the electron is present.

Hence difference between $n_2,n_1$ orbit's energy $\Delta E=E_{n_2}-E_{n_1} \propto (\frac{1}{n_2^2}-\frac{1}{n_1^2})$ $\Delta E=\frac{hc}{\lambda} \propto (\frac{1}{n_2^2}-\frac{1}{n_1^2})$ $\frac{1}{\lambda} \propto (\frac{1}{n_2^2}-\frac{1}{n_1^2})$ Now taking $\lambda_1=12818 angstrom$ $\frac{\lambda_2}{\lambda_1}=\frac{\frac{1}{3^2}-\frac{1}{5^2}}{\frac{1}{1^2}-\frac{1}{3^2}}$

$\lambda_2= \lambda_1 \times \frac{2}{25}$ $\lambda_2=102.5 nm$

Note:Actually , $\Delta E=E_{n_2}-E_{n_1} = IE(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ where $IE$ is ionization of single electron species.

Moderator note:

Great solution. Does any of this generalize to more complicated atoms?

You obtained 102.544 nm. There may be people who suggest for 102.6 nm.

Lu Chee Ket - 5 years, 3 months ago

+1 ! Nice Sol $^n$

Rajdeep Dhingra - 5 years, 10 months ago

Modern Physics or Chemistry?

The answer is 102.5.

1 solution

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