Question

The energy of electron in hydrogen atom is given by $E_{n}=-21 \cdot 7 \times 10^{-12}$ erg per atom/n $n^{2}$. Calculate the energy required to remove an electron completely from $n=2$ orbit. What is the longest wavelength (in $cm$ ) of light that can be used to cause this transition?

Answer 1

SOLUTION : Given, $E_{n}=-\frac{21 \cdot 7 \times 10^{-12}}{n^{2}}$ erg per atom

$\therefore \quad E_{2}=-\frac{21.7 \times 10^{-12}}{2^{2}}$ erg per atom and $E_{\infty}=-\frac{21 \cdot 7 \times 10^{-12}}{\infty^{2}}$ erg $=0$

$\therefore \quad E_{\infty}-E_{2}=\frac{21 \cdot 7 \times 10^{-12}}{4}$ erg per atom $=5 \cdot 425 \times 10^{-12}$ erg per atom.

Second part :

$ E=h v=\frac{h c}{\lambda} \quad \therefore \quad \lambda=\frac{h c}{E}=\frac{6 \cdot 62 \times 10^{-27} erg . sec \times 3 \times 10^{10} cm s ^{-1}}{5 \cdot 425 \times 10^{-12} erg } \\\therefore \quad \lambda= 3.66 \times 10^{-5} cm =3660 Å $