Question

Bohr Model: an electron in a doubly ionized lithium atom +2Li(three protons in the nucleus) makes a transition from the 풏=ퟏto the 풏=ퟑlevel with an associated photon. a)Determine the photon energy associated with this transition. (10pts)

Answer 1

Answer:

$\lambda =113.74{\AA}$

Explanation:

Electron is jumping from n=1 to n=3

and for Li Z=3;

$E_n=-13.6\times \frac{Z^2}{n^2}$

$\Delta E=-13.6\times Z^2[\frac{1}{n_3^2} -\frac{1}{n_1^2}]$

$n_3=3 \,and \,n_1=1$

after solving we get;

$\Delta E=-13.6\times 8 \,ev$

for the photon energy is calculated by the following equation

$\Delta E=\frac{hc}{\lambda}$

for easy calculation;

$\Delta E=\frac{12375}{\lambda} {\AA}$

where energy is taken in electron volt

after puting value of $\Delta E$ we get,

$\lambda =113.74{\AA}$