Question

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).

Answer 1

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:

$E_0=-(2.18\times {10}^{-18})\frac{Z^2}{n^2}$

Where:

$(Z)$ is the atomic number of the atom.

$(n)$ is the principal quantum number.

To determine the energy required for ionization, transitioning from $(n_1=5)to(n_2=\mathrm{\infty })$:

$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_5$

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Using the given formula:

$\mathrm{\Delta }E=\{-(2.18\times {10}^{-18}\text{J})\frac{1^2}{{\mathrm{\infty }}^2}\}-\{-(2.18\times {10}^{-18}\text{J})\frac{1^2}{5^2}\}$

Given that $(\frac{1}{\mathrm{\infty }})$ approximates to 0, the equation simplifies to:

$$\mathrm{\Delta }E=(2.18\times {10}^{-18}\text{J})\frac{1}{25}$$

$$\mathrm{\Delta }E=8.72\times {10}^{-20}\text{J}$$

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 $(n_1=1)to(n_2=\mathrm{\infty })$:

$\mathrm{\Delta }E=E_{\mathrm{\infty }}-E_1$

By substituting in the given values:

$\mathrm{\Delta }E=(2.18\times {10}^{-18})=2.18\times {10}^{-18}\text{J}$

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.