Question

Calculate the ratio of KE in second and fourth excited state of hydrogen atom.

Answer 1

Kinetic Energy of a Bohr's Model of a atom has a formula.

$E=\frac{-13.6\left(ev\right)z^2}{n^2}$

where , $z\to$ atomic number.

$n\to$ principal Quantum number.

For second excited state, $n=3$

For fourth excited state , $n=5$

Ratio $=\frac{-13.6}{9}\times \frac{25}{-13.6}=\frac{25}{9}$

$=25:9$