An electronic transition in hydrogen atom results in the formation of $H_{\alpha}$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $mol^{- 1}$) are

Question

Dr. Abubacker · Accepted Answer

(b)
Energy of an electron in $n$th orbit,
$E_{n} = \frac{2\pi^{2}k^{2}mZ^{2}e^{4}}{n^{2}h^{2}}$
On submitting the values of $k,\ m,\ e$ and $h$, we get
$E_{n} = - \frac{2.172 	imes 10^{- 18}Z^{2}}{n^{2}}J\ atom^{- 1}$
$or\  = - \frac{1311.8Z^{2}}{n^{2}}kJ\ mol^{- 1}$
$or = - \frac{313.52Z^{2}}{n^{2}}kcal\ mol^{- 1}\lbrack\because 1\ kcal = 4.184\ kJbrack$
For H-atom, $Z = 1$
For Lyman series, $n_{1} = 1,\ n_{2} = 2$
Energy of electron in $n_{1}$ orbit
$= - \frac{313.52 	imes (1)^{2}}{(1)^{2}}kcal\ mol^{- 1}$
$= - 313.52\ kcal\ mol^{- 1}$
$\approx - 313.6\ kcal\ mol^{- 1}$
Energy of electron in $n_{2}$ orbit
$= - \frac{313.52 	imes (1)^{2}}{(2)^{2}}kcal\ mol^{- 1}$
$= - \frac{313.52}{4}kcal\ mol^{- 1}$
$= - 78.38\ kcal\ mol^{- 1}$

Q. An electronic transition in hydrogen atom results in the formation of $H_{\alpha}$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $mol^{- 1}$) are

Explanation

Reviewed by Dr. Abubacker

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