Question

Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The energy needed to ionize a ground-state hydrogen atom is 2.18 x 10–18 J.

Answer 1

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

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

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

$E_{ionization}=5.45\times 10^{-19}\ J$

Answer 2

The ionization energy required to ionize a hydrogen atom from n = 2 energy level is 5.45 × 10⁻¹⁹

The energy required for the ionization of an atom is the amount of energy needed to remove an electron from an atom from its excited state to a ground state.

The ionization energy can be expressed by using the formula:

$\mathbf{E= 2.18 \times 10^{-18} \Big (\dfrac{1}{n_1^2} - \dfrac{1}{\infty^2} \Big)}$

where;

n = 2

$\mathbf{E= 2.18 \times 10^{-18} \Big (\dfrac{1}{2^2} - \dfrac{1}{\infty^2} \Big)}$

$\mathbf{E=\dfrac{ 2.18 \times 10^{-18}}{4} }$

$\mathbf{E=5.45 \times 10^{-19}}$

Learn more about ionization energy here:

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