# 11. An electron jumps from 3rd orbit to 1st orbit in H-atom. Find out the wavelength of emitted radiations?

Jun 24, 2018

$\lambda = 102.6 \textcolor{w h i t e}{l} n m$

#### Explanation:

The energy of the photon emitted shall equal to the difference between the electron's potential energy at the two orbits.

The $\underline{\text{Rydberg Formula}}$ suggests that an electron that travels (or literally "falls") from energy level ${n}_{i}$ to ${n}_{f}$ (${n}_{i} > {n}_{f}$) emits a photon of wavelength $\lambda$ for which

$\frac{1}{\lambda} = R \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

where the Rydberg's Constant $R = 1.097 \times {10}^{7} \textcolor{w h i t e}{l} {m}^{- 1}$.

In this particular scenario

• ${n}_{i} = 3$
• ${n}_{f} = 1$

Such that

$\frac{1}{\lambda} = R \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$
$\textcolor{w h i t e}{\Delta \text{PE}} = 1.097 \times {10}^{7} \textcolor{w h i t e}{l} {m}^{- 1} \cdot \left(\frac{1}{1} ^ 2 - \frac{1}{3} ^ 2\right)$
$\textcolor{w h i t e}{\Delta \text{PE}} = 9.751 \cdot {10}^{7} \textcolor{w h i t e}{l} {m}^{- 1}$

$\lambda = 1.026 \times {10}^{- 7} \textcolor{w h i t e}{l} \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot \frac{{10}^{9} \textcolor{w h i t e}{l} n m}{1 \textcolor{w h i t e}{l} \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}}}$
$\textcolor{w h i t e}{\lambda} = 102.6 \textcolor{w h i t e}{l} n m$

References
"Bohr's Hydrogen Atom" , Chemistry Libretext