
The ionization energy of a hydrogen atom is $ 13.6eV $ . The energy required to excite the electron in a hydrogen atom from the ground state to its first excited state is:
(A) $ 1.69 \times {10^{ - 18}}J $
(B) $ 1.69 \times {10^{ - 23}}J $
(C) $ 1.69 \times {10^{23}}J $
(D) $ 1.69 \times {10^{25}}J $
Answer
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Hint :Ground state is the lowest energy state of an atom. On the other hand, excited state is the state when an atom is at a higher energy level than at the ground state. Ionization energy is the energy required to remove the lowest orbiting electron from the influence of the central protons.
Complete Step By Step Answer:
It is given that the ionization energy of a hydrogen atom is $ 13.6eV $ . We know that the transition energy for ionization process from the state $ {n_i} $ to next state $ {n_f} $ is given by,
$ \Delta E = {E_f} - {E_i} $ or $ \Delta E = {E_2} - {E_1} $
Now, it is given that $ {E_1} = 13.6eV $ . For calculating $ {E_2} $ , we will use the formula $ {E_n} = \dfrac{{{E_1}}}{{{n^2}}} $ . Here, $ n = 2 $ therefore, $ {E_2} = \dfrac{{ - 13.6}}{{{2^2}}} = - 3.4eV $ .
Transition energy from ground to first excited state is:
$ \Delta E = {E_2} - {E_1} $
$ \Delta E = - 3.4 - ( - 13.6) $
$ \Delta E = 10.2eV $
Now, we need to convert the above result into joules. We know that $ 1eV = 1.602 \times {10^{ - 19}}J $ .
Hence, $ 10.2eV = 10.2 \times 1.602 \times {10^{ - 19}}J = 1.63 \times {10^{ - 18}}J $ .
This result corresponds to the option A.
Hence, the correct option for the given question is A.
Note :
In ground state, all electrons are in the lowest possible energy levels and the total energy of the electrons cannot be lowered by transferring one or more electrons to different orbitals. Excitation of electrons happens due to absorption of energy by them.
When we move across the periodic table, the ionization energy increases from left to right. This happens because the atomic number of elements increases which means that the number of protons and electrons in the atoms increases. Due to the large positive charge on the nucleus, the valence electrons are pulled in more strongly by the nucleus and it becomes more and more difficult for the atoms to lose electrons. Thus, on moving from left to right in a period, the tendency of atoms to lose electrons decreases. Hence, the ionization energy increases across the period. Similarly, the ionization energy decreases on moving down a group.
Complete Step By Step Answer:
It is given that the ionization energy of a hydrogen atom is $ 13.6eV $ . We know that the transition energy for ionization process from the state $ {n_i} $ to next state $ {n_f} $ is given by,
$ \Delta E = {E_f} - {E_i} $ or $ \Delta E = {E_2} - {E_1} $
Now, it is given that $ {E_1} = 13.6eV $ . For calculating $ {E_2} $ , we will use the formula $ {E_n} = \dfrac{{{E_1}}}{{{n^2}}} $ . Here, $ n = 2 $ therefore, $ {E_2} = \dfrac{{ - 13.6}}{{{2^2}}} = - 3.4eV $ .
Transition energy from ground to first excited state is:
$ \Delta E = {E_2} - {E_1} $
$ \Delta E = - 3.4 - ( - 13.6) $
$ \Delta E = 10.2eV $
Now, we need to convert the above result into joules. We know that $ 1eV = 1.602 \times {10^{ - 19}}J $ .
Hence, $ 10.2eV = 10.2 \times 1.602 \times {10^{ - 19}}J = 1.63 \times {10^{ - 18}}J $ .
This result corresponds to the option A.
Hence, the correct option for the given question is A.
Note :
In ground state, all electrons are in the lowest possible energy levels and the total energy of the electrons cannot be lowered by transferring one or more electrons to different orbitals. Excitation of electrons happens due to absorption of energy by them.
When we move across the periodic table, the ionization energy increases from left to right. This happens because the atomic number of elements increases which means that the number of protons and electrons in the atoms increases. Due to the large positive charge on the nucleus, the valence electrons are pulled in more strongly by the nucleus and it becomes more and more difficult for the atoms to lose electrons. Thus, on moving from left to right in a period, the tendency of atoms to lose electrons decreases. Hence, the ionization energy increases across the period. Similarly, the ionization energy decreases on moving down a group.
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