VerifiedHint: Electron's ionization energy decreases with increasing principal quantum number. Compare using the formula E = -13.6 $\times \frac{Z^2}{n^2}$
Step by Step Solution:
The formula representing the energy of an electron in an orbit of a hydrogen atom is:
$ E_0 = - \left(2.18 \times 10^{-18}\right) \frac{Z^2}{n^2} $
Where:
$( Z )$ is the atomic number of the atom.
$( n )$ is the principal quantum number.
To determine the energy required for ionization, transitioning from $( n_1 = 5 )\ to\ ( n_2 = \infty )$:
$ \Delta E = E_\infty - E_5 $
Using the given formula:
$ \Delta E = \left\{ -\left(2.18 \times 10^{-18} \text{J}\right) \frac{1^2}{\infty^2} \right\} - \left\{ -\left(2.18 \times 10^{-18} \text{J}\right) \frac{1^2}{5^2} \right\}$
Given that $( \frac{1}{\infty} )$ approximates to 0, the equation simplifies to:
\[ \Delta E = \left(2.18 \times 10^{-18} \text{J}\right) \frac{1}{25} \]
\[ \Delta E = 8.72 \times 10^{-20} \text{J} \]
Thus, the energy needed to ionize an electron from the n=5 orbit to infinity is
$( 8.72 \times 10^{-20} \text{J} )$.
For ionization from $( n_1 = 1 ) to ( n_2 = \infty )$:
$ \Delta E = E_\infty - E_1 $
By substituting in the given values:
$ \Delta E = \left(2.18 \times 10^{-18}\right) = 2.18 \times 10^{-18} \text{J} $
In conclusion, ionizing an electron in the 5th orbit of a hydrogen atom requires less energy compared to ionizing one in the ground state.
Note: The formula E = -13.6 $\times \left(\frac{Z^2}{n^2}\right)$ represents the energy levels of electrons in an atom. In this equation, E denotes the energy, Z represents the atomic number, and n signifies the principal quantum number. This formula, derived from the Bohr model, calculates the energy associated with each electron's orbit within an atom.
