Question

If the binding energy of 2nd excited state of
hydrogen like sample is 24 eV approximately, then the ionization energy of the sample is​

Answer 1

Answer:

19.14 eV the ionization energy of the sample.

Explanation:

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

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Second excited state = 1 → 3

$24eV=E_{2}-E_1$

$24eV=(-13.6\times \frac{Z^2}{3^2}eV)-(-13.6\times \frac{Z^2}{1^2}eV)$

Z = 1.408 amu

Energy required to remove to ionize the atom.

1 → ∞

$E=(-13.6\times \frac{(1.408)^2}{\infty ^2}eV)-(-13.6\times \frac{(1.408)^2}{1^2}eV)$

$=19.14 eV$

19.14 eV the ionization energy of the sample.