Question

Find the wavelength of the radiation required to excite the electron in $L{i}^{++}$ from the first to the third Bohr orbit. You may need to use the constant hc = 1242 eV.nm.

• A. 13.6 nm
• B. 11.4 nm
• C. 27.2 nm
• D. 6.8 nm

Answer 1

The correct option is B 11.4 nm

So I have to give sufficient energy to excite an electron from the $1^{st}$ orbit of $L{i}^{++}$ ion to $3^{rd}$. So what is the energy of $L{i}^{++}$ ion in the ground state.
$E_1=\frac{-13.6Z^2}{n^2}$.
Z = 3, n = 1

$E_1$ = 9 $\times$ (-13.6) eV.
Similarly if I give it sufficient energy and the electron reaches 3rd orbit then the energy of the ion will be

$E_2=\frac{-13.6Z^2}{n^2}$.
Z = 3, n = 3

$E_2=-13.6eV.$

So the energy that would be required to achieve it should be.
$E_2-E_1=8\times \left(13.6\right)=108.8eV.$

Now the wavelength of this light of energy 108.8 eV could be found by .....
I know $\frac{hc}{\lambda }$.
Simple

$\frac{hc}{\lambda }$ = 108.8

$\lambda$ = 11.4 nm