QJEEM23Ch039a: JEE Main 2023 Chemistry - Class 11 JEEM23Ch039a Solution: The energy of an electron in the first Bohr orbit of hydrogen…

The energy of an electron in a hydrogen atom is given by the following formula:

$E_n = \frac{-13.6\, \mathrm{eV}}{n^2}$

where n is the principal quantum number (Bohr orbit number). To find the energy in the third Bohr orbit (n = 3), we can use the formula:

$E_3 = \frac{-13.6\, \mathrm{eV}}{3^2} = \frac{-13.6\, \mathrm{eV}}{9}$

Now, we are given the energy of the electron in the first Bohr orbit (n = 1):

$E_1 = -2.18 \times 10^{-18}\, \mathrm{J}$

First, let's convert this energy to electron volts (eV):

$E_1 = -2.18 \times 10^{-18}\, \mathrm{J} \times \frac{1\, \mathrm{eV}}{1.6 \times 10^{-19}\, \mathrm{J}} \approx -13.6\, \mathrm{eV}$

Now we can find the ratio of $E_3$ to $E_1$ :

$\text{Ratio} = \frac{E_3}{E_1} = \frac{-\frac{13.6\, \mathrm{eV}}{9}}{-13.6\, \mathrm{eV}} = \frac{1}{9}$

Thus, the energy of the electron in the third Bohr orbit is $\frac{1}{9}$ th of its energy in the first Bohr orbit.