Atomic StructureHard
Question
The amount of energy released when an electron jumps from the seventh excited state to the first excited state in He+ ion is
Options
A.13.32 eV
B.53.28 eV
C.12.75 eV
D.26.08 eV
Solution
$\Delta E = 13.6z^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)eV = 13.6 \times 2^{2}\left( \frac{1}{2^{2}} - \frac{1}{8^{2}} \right) = 12.75\text{ eV}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The ratio of the speed of the electron in the ground state of hydrogen atom to the speed of light in vacuum is...Which of the following pairs of ions are isoelectronic and isostructural ?...Neon does not form clathrate compound because -...Identify the INCORRECT statements from the following:A. Notation $\ _{12}^{24}Mg$ represents 24 protons and 12 neutrons....The de Broglie wavelength of a vehicle moving with velocity v is λ. Its load is changed so that the velocity as well as ...