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If the ionization energy of hydrogen atom is $13.6$ eV, the energy required to excite it from ground state to the next higher state is approximately:
A) $3.4$ eV
B) $10.2$ eV
C) $17.2$ eV
D) $13.6$ eV

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Last updated date: 20th Jun 2024
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Answer
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Hint:We know that ground state is the lowest energy state that an atom can have. This is the energy state that is considered normal for an atom. On the other hand, the excited state is that state where an atom is at a higher energy level than the ground state. When the atoms or electrons absorb energy, they jump to the higher orbital and become excited.

Complete step-by-step answer:We know that for ionization process transition is from ${n_i} = 1$ to ${n_f} = \infty ,$
Also, ionization energy $ = 13.6$ eV which is already given in the question.
So, the Transition energy for ionization process will be:
$\Delta E = {E_\infty } - {E_1} = IE$
as we know that ${E_\infty } = 0$
thus ${E_1} = - 13.6eV$
Since the first excited state is $n = 2$
therefore, by applying ${E_n} = \dfrac{{{E_1}}}{{{n^2}}}$ as a formula we will calculate the value of ${E_2}$
so, let’s calculate the value of ${E_2}$
${E_2} = \dfrac{{ - 13.6}}{{{2^2}}}$
${E_2} = - 3.4eV$
Transition energy from ground state i.e., $n = 2$ in Hydrogen atom is given as:
$\Delta E = {E_2} - {E_1}$
$\Delta E = - 3.4 - ( - 13.6)$
$\Delta E = 10.2eV$
Therefore, If the ionization energy of hydrogen atom is $13.6$ eV, the energy required to excite it from ground state to the next higher state will be equal to $10.2$ eV.

Hence the correct answer is Option B).

Note:Ionization energy is the energy required to remove the lowest orbiting electron from the influence of the central protons. while moving across the period (left to right) the ionization energy increases and it decreases on moving down a group (top to bottom).