
The energy of the electron in the hydrogen atom is given by the expression atom in the ground state can absorb is:
A.\[13.16{\text{ eV}}\]
B.\[32.8{\text{ eV}}\]
C.\[65.6{\text{ eV}}\]
D.\[984\,{\text{eV}}\]
Answer
557.1k+ views
Hint:To answer this question, you should recall the energy of orbitals. The energy of orbitals refers to the energy required to take an electron present in that orbital to infinity or the energy released when an electron is added to that orbital from infinity. It is inversely proportional to the square of principal quantum number.
Complete step by step answer:
The exception to the general behaviour of the energy of orbitals is observed in Hydrogen, the energy of the orbital is only dependent on principal quantum number, and so the 2s and 2p orbital in hydrogen atom have the same energy. The 1s orbital in hydrogen atom corresponds to the most stable condition and is called ground state whereas any other orbital afterwards has higher energy than that of 1s orbital and is called an excited state.
The energy of an electron is given by, \[{E_n} = \dfrac{{13.16Z}}{{{n^2}}}eV\].
Here, Z is the atomic mass number and n is the principal quantum number. For the ground state of a hydrogen atom \[Z = 1\] and \[n = 1\;\], \[E = 13.16{\text{ eV}}\].
Hence, the correct answer is option A.
Note:
An electron absorbs energy in the form of photons and gets excited to a higher energy level. After jumping to the higher energy level, also called the excited state, the excited electron is less stable, and therefore, would quickly emit a photon to come back to a lower and more stable energy level. The emitted energy is equivalent to the difference in energy between the two energy levels for a specific transition. This phenomenon occurs by the gain of energy where an electron moves from a higher energy level to lower energy level by losing energy. This energy is lost in the form of spectral lines. Hence spectral series can be defined as a set of wavelengths arranged sequentially with the characteristic feature of every atom. The emitted wavelengths are resolved using a spectroscope.
Complete step by step answer:
The exception to the general behaviour of the energy of orbitals is observed in Hydrogen, the energy of the orbital is only dependent on principal quantum number, and so the 2s and 2p orbital in hydrogen atom have the same energy. The 1s orbital in hydrogen atom corresponds to the most stable condition and is called ground state whereas any other orbital afterwards has higher energy than that of 1s orbital and is called an excited state.
The energy of an electron is given by, \[{E_n} = \dfrac{{13.16Z}}{{{n^2}}}eV\].
Here, Z is the atomic mass number and n is the principal quantum number. For the ground state of a hydrogen atom \[Z = 1\] and \[n = 1\;\], \[E = 13.16{\text{ eV}}\].
Hence, the correct answer is option A.
Note:
An electron absorbs energy in the form of photons and gets excited to a higher energy level. After jumping to the higher energy level, also called the excited state, the excited electron is less stable, and therefore, would quickly emit a photon to come back to a lower and more stable energy level. The emitted energy is equivalent to the difference in energy between the two energy levels for a specific transition. This phenomenon occurs by the gain of energy where an electron moves from a higher energy level to lower energy level by losing energy. This energy is lost in the form of spectral lines. Hence spectral series can be defined as a set of wavelengths arranged sequentially with the characteristic feature of every atom. The emitted wavelengths are resolved using a spectroscope.
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