# Estimate the ratio of the wavelengths associated with the electron orbiting around the nucleus in the first and second excited states of hydrogen atom.

The wavelength can find by using De-Broglie wavelength equation,

for electron, the first excited n=2.

By de-broglie's equation, $n\lambda =2\pi r _{n}$

where, $r _{n}$ is the radius of second orbit,

$r _{n}=n^{2}\left ( \frac{h^{2}\epsilon _{0}}{\pi m_{e}^{2}} \right )$

$=n^{2}\times 0.527\; A^{\circ}$

$=2.11\; A^{\circ}$

Therefore, $2\times \lambda =2\times 3.14\times 2.11$

$\lambda =6.625\; A^{\circ}$

For second excited state, n=3

So, $n\lambda =2\pi r_{n}$

$r_{n}=n^{2}\times 0.529\; A^{\circ}$

$=9\times 0.529\; A^{\circ}=4.761\; A^{\circ}$

Therefore, $3\times \lambda =2\times 3.14\times 4.761$

$\lambda =\frac{29.899}{3}=9.966\; A^{\circ}$

Ratio,               $=\frac{6.625}{9.966}$

$=0.66$

ratio of the wave lengths is $0.66$

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