Question

if the electron fall from n=5 to n=4 in h atom then emitted energy is

Answer 1

∆E=13.6×Z^2(1/4^2-1/5^2)
∆E=13.6×1(1/16-1/25)
∆E=13.6(9/400)
∆E=1.278

Answer 2

Answer:

If the electron falls from n = 5 to n = 4 in H-atom then emitted energy will be equal to 0.306eV.

Explanation:

We know that the energy of nth energy level of H-atom is given by:

$E=-13.6\frac{Z^2}{n^2}\; eV$

Where Z is an atomic number, for H-atom, Z=1

Therefore, $E=-\frac{13.6}{n^2}\; eV$

The energy of n = 5, fifth energy level of H-atom:

$E_{5}=-\frac{13.6}{(5)^2}\; eV$

$E_{5}=-0.544eV$

The energy of n=4, fourth energy level of hydrogen atom:

$E_{4}=-\frac{13.6}{(4)^2}\; eV$

$E_{4}=-0.84eV$

When an electron fall from higher energy level to lower energy level, then energy will be emitted.

Energy emitted = E₅ - E₄ $=-0.544-(-0.85)$

Emitted energy = 0.306eV

Therefore the energy emitted in the given transition is equal to 0.306eV.