The energy of an electron in the second orbit of a hydrogen atom is $ -3.4 \, \text{eV} $. What is the energy of the electron in the third orbit? (Given: Energy of an electron in the $ n $-th orbit of hydrogen is $ E_n = -\frac{13.6}{n^2} \, \text{eV} $).

Question

Two electrons are moving in orbits of two hydrogen like atoms with speeds $3\times10^{5}\,\text{m/s}$ and $2.5\times10^{5}\,\text{m/s}$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are ___ respectively.

Answer 1

The energy of an electron in the $ n $-th orbit of a hydrogen atom is $ E_n = -\frac{13.6}{n^2} \, \text{eV} $. For the third orbit ($ n = 3 $): $ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \, \text{eV} $. This is approximately $ -1.511 \, \text{eV} $. Rounded to two decimal places, $ E_3 \approx -1.51 \, \text{eV} $. For verification, the energy in the second orbit ($ n = 2 $) is $ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} $. This matches the given value, showing the formula's consistency. As the orbit number increases, the energy becomes less negative (higher). Therefore, the third orbit has a higher energy than the second. The energy of the electron in the third orbit is $ -1.51 \, \text{eV} $.

The energy of an electron in the second orbit of a hydrogen atom is \( -3.4 \, \text{eV} \). What is the energy of the electron in the third orbit? (Given: Energy of an electron in the \( n \)-th orbit of hydrogen is \( E_n = -\frac{13.6}{n^2} \, \text{eV} \)).

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on Atomic Models

Questions Asked in MHT CET exam