Question.
The electron energy in hydrogen atom is given by $E_{n}=\left(-2.18 \times 10^{-18}\right) / n^{2} \mathrm{~J}$. Calculate the
energy required to remove an electron completely from the n = 2 orbit. What is the longest
wavelength of light in cm that can be used to cause this transition?
The electron energy in hydrogen atom is given by $E_{n}=\left(-2.18 \times 10^{-18}\right) / n^{2} \mathrm{~J}$. Calculate the
energy required to remove an electron completely from the n = 2 orbit. What is the longest
wavelength of light in cm that can be used to cause this transition?
Solution:
Given
$E_{n}=-\frac{2.18 \times 10^{-18}}{n^{2}} \mathrm{~J}$
Energy required for ionization from n = 2 is given by,
$\Delta E=E_{\infty}-E_{2}$
$=\left[\left(\frac{-2.18 \times 10^{-18}}{(\infty)^{2}}\right)-\left(\frac{-2.18 \times 10^{-18}}{(2)^{2}}\right)\right] J$
$=\left[\frac{2.18 \times 10^{-18}}{4}-0\right] \mathrm{J}$
$=0.545 \times 10^{-18} \mathrm{~J}$
$\Delta \mathrm{E}=5.45 \times 10^{-19} \mathrm{~J}$
$\lambda=\frac{\mathrm{hc}}{\Delta E}$
Here, $\lambda$ is the longest wavelength causing the transition.
$\lambda=\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{5.45 \times 10^{-19}}=3.647 \times 10^{-7} \mathrm{~m}$
$=3647 \times 10^{-10} \mathrm{~m}$
$=3647 \AA$
Given
$E_{n}=-\frac{2.18 \times 10^{-18}}{n^{2}} \mathrm{~J}$
Energy required for ionization from n = 2 is given by,
$\Delta E=E_{\infty}-E_{2}$
$=\left[\left(\frac{-2.18 \times 10^{-18}}{(\infty)^{2}}\right)-\left(\frac{-2.18 \times 10^{-18}}{(2)^{2}}\right)\right] J$
$=\left[\frac{2.18 \times 10^{-18}}{4}-0\right] \mathrm{J}$
$=0.545 \times 10^{-18} \mathrm{~J}$
$\Delta \mathrm{E}=5.45 \times 10^{-19} \mathrm{~J}$
$\lambda=\frac{\mathrm{hc}}{\Delta E}$
Here, $\lambda$ is the longest wavelength causing the transition.
$\lambda=\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{5.45 \times 10^{-19}}=3.647 \times 10^{-7} \mathrm{~m}$
$=3647 \times 10^{-10} \mathrm{~m}$
$=3647 \AA$