The electron energy in hydrogen atom is given by $\mathrm{E}=\left(-21.7 imes 10^{-12}ight) / \mathrm{n}^{2}$ ergs. By calculating the energy required to remove an electron completely from the $\mathrm{n}=2$ orbit, find out the longest wavelength (in $\mathrm{cm}$ ) of light that can be used to…

Question

The electron energy in hydrogen atom is given by $\mathrm{E}=\left(-21.7 	imes 10^{-12}ight) / \mathrm{n}^{2}$ ergs. By calculating the energy required to remove an electron completely from the $\mathrm{n}=2$ orbit, find out the longest wavelength (in $\mathrm{cm}$ ) of light that can be used to cause this transition?

Aicharya · Accepted Answer

$\Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_{2}=[0]-\left[-\frac{21.7 \times 10^{-12}}{4}ight]$
Since $E_{\mathrm{n}}=-\frac{21.7 \times 10^{-12}}{\mathrm{n}^{2}}$ ergs $=\frac{21.7 \times 10^{-12}}{4} \quad \Delta \mathrm{E}=\mathrm{hv}$
$\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \quad \frac{21.7 \times 10^{-12}}{4}=6.627 \times 10^{-27} \times \frac{3 \times 10^{10}}{\lambda}$
whence $\lambda=3.67 \times 10^{-5} \mathrm{~cm}$ ,

The electron energy in hydrogen atom is given by $\mathrm{E}=\left(-21.7 \times 10^{-12}ight) /…

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