Question

The energy of the second orbit of hydrogen is equal to the energy of:
A. Fourth orbit of $H{e^ + }$
B. Fourth orbit of $L{i^{2 + }}$
C. Second orbit of $H{e^ + }$
D. Second orbit of $L{i^{2 + }}$

Answer 1

Hint: The energy of the second Bohr orbit of the hydrogen atom is $ - 328kJ/mol.$ and the energy of $H$- atom is inversely proportional to the square of principle quantum number the energy of fourth Bohr.

Complete step by step answer:
The Bohr model was introduced by Niels Bohr in 1913. It is a system consisting of a small, dense nucleus surrounded by orbiting electrons. It is similar to the structure of the solar system but it provides attraction of electrostatic forces instead of gravity. The atom will be completely stable in the state with the smallest orbit after that there is no orbit of lower energy into which the electron can jump. According to the question –

$E = - \dfrac{Z^2}{n^2} \times 13.6eV$

${E_2} = - \dfrac{13.6}{4}$ for ‘H’

By using the formula,

$\Rightarrow E = - \dfrac{Z^2}{n^2} \times 13.6eV$

$\Rightarrow - \dfrac{13.6}{4} = - \dfrac{Z^2}{n^2} \times 13.6$

$\Rightarrow \dfrac{Z^2}{n^2} = \dfrac{1}{4}$ ($Z = 2$, $n = 4$)

$Z$ is represented by an atomic number.

The Quantum number is represented by $n$. Therefore, The energy of the second orbit of hydrogen is equal to the energy of the Fourth orbit of $H{e^ + }$.

So, the correct answer is Option A.

Note: In the Bohr model that electrons propagate in defined circular orbits around the nucleus. The quantum number (n) and an integer are used to label the orbits and electrons can jump from one orbit to another by absorbing or emitting the energy.

As we move away from the nucleus, the energy of the orbit increases. Bohr’s orbit is a hypothetical path of an electron about the nucleus of the Bohr atom.