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# Question:How much energy is required to ionize an H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of the H atom (energy required to remove the electron from n =1 orbit).

Last updated date: 20th Jun 2024
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Hint: 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.