Question

2nd excited energy for hydrogen like atom is 24ev find ionization energy

Answer 1

Answer:

Sutar

Explanation:

Answer 2

Answer:

19.14 eV the ionization energy of the sample.

Explanation:

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

n

=−13.6×

n

2

Z

2

ev

where,

E_nE

n

= energy of n^{th}n

th

orbit

n = number of orbit

Z = atomic number

Second excited state = 1 → 3

24eV=E_{2}-E_124eV=E

2

−E

1

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

3

2

Z

2

eV)−(−13.6×

1

2

Z

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)E=(−13.6×

∞

2

(1.408)

2

eV)−(−13.6×

1

2

(1.408)

2

eV)

=19.14 eV=19.14eV

19.14 eV the ionization energy of the sample.