Answer:
[tex]\lambda =113.74{\AA}[/tex]
Explanation:
Electron is jumping from n=1 to n=3
and for Li Z=3;
[tex]E_n=-13.6\times \frac{Z^2}{n^2}[/tex]
[tex]\Delta E=-13.6\times Z^2[\frac{1}{n_3^2} -\frac{1}{n_1^2}][/tex]
[tex]n_3=3 \,and \,n_1=1[/tex]
after solving we get;
[tex]\Delta E=-13.6\times 8 \,ev[/tex]
for the photon energy is calculated by the following equation
[tex]\Delta E=\frac{hc}{\lambda}[/tex]
for easy calculation;
[tex]\Delta E=\frac{12375}{\lambda} {\AA}[/tex]
where energy is taken in electron volt
after puting value of [tex]\Delta E[/tex] we get,
[tex]\lambda =113.74{\AA}[/tex]
