Answer :
Answer:
(a) 0.3778 eV
(b) Ratio = 0.0278
Explanation:
The Bohr's formula for the calculation of the energy of the electron in nth orbit is:
[tex]E=\frac {-13.6}{n^2}\ eV[/tex]
(a) The energy of the electron in n= 6 excited state is:
[tex]E=\frac {-13.6}{6^2}\ eV[/tex]
[tex]E=-0.3778\ eV[/tex]
Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV
(b) For first orbit energy is:
[tex]E=\frac {-13.6}{1^2}\ eV[/tex]
[tex]E=-13.6\ eV[/tex]
[tex]Ratio=\frac {E_6}{E_1}[/tex]
[tex]Ratio=\frac {-0.3778}{-13.6}[/tex]
Ratio = 0.0278