Question

The ionization energy of a hydrogen-like Bohr atom in the ground state is 4 rydberg.

(a) What is the wavelength of the radiation emitted when the electron jumps from the second orbit to first orbit?

(b) What is the radius of the first orbit of the atom? 1 rydberg $=13.6 eV )$.

Answer 1

(a) The l.E. of the atom $=4 \times 13.6 eV =54.4 eV$

Therefore, the energy of the first orbit $=-54.4 eV$

$\because \quad E_{n}=-13 \cdot 6 \times \frac{Z^{2}}{n^{2}} cV$

$\quad \therefore-54 \cdot 4=-13 \cdot 6 \times \frac{Z^{2}}{1^{2}}$

$\quad \therefore \quad Z=2$

The atomic number of the $H$ - like atom $=2$

For $H$ - like atom,

$\Delta E=E_{2}-E_{1}=13 \cdot 6 \times Z^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) eV$

$=13 \cdot 6 \times 4 \times \frac{3}{4} eV =40 \cdot 8 eV \\\lambda=\frac{h c}{\Delta E}=\frac{6 \cdot 626 \times 10^{-34} \times 3 \times 10^{8}}{40 \cdot 8 \times 1 \cdot 60 \times 10^{-19}} m$

$=0 \cdot 304 \times 10^{-7} m =30 \cdot 4 nm$

(b) $r_{n}=0.529 \times \frac{n^{2}}{Z} Å=\frac{0.529 \times 1}{2} Å=0.2645 Å $.