Question

calculate the energy values of the electron in second and third orbits of hydrogen atom

Answer 1

Answer:

Explanation:

The energy of an electron in the second orbit of hydrogen  atom is 5.4475 × 10⁻²² j

The radius of the second orbit : 2.1168 $E_{n} = 2.179$ 10⁻¹⁰m

The energy on the nth orbital is given by = $E_{n} = k\frac{z^{2} }{n^{2} }$

Where  k = 2.179 × 10⁻¹⁸ j

The energy of an electron in the second orbit of hydrogen  atom :  n = 2, Z=1

$E_{n} = 2.179$ × 10⁻¹⁸ j $\frac{1^{2} }{2^{2} } = 5.4475$ ×10⁻²² j

The radius of the nth orbit is given by : $r_{n} = \frac{n^{2} }{z}$×$a_{o}$

$a_{o} =$ 5.292 × 10⁻¹¹$m$

The radius of the second orbit :

$r_{2} = \frac{2^{2} }{1}$ × 5.292 × 10⁻¹¹$m$ = 2.1168 × 10⁻¹⁰$m$