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Question: A single electron orbits a stationary nucleus \( \left( {Z = 5} \right) \) .The energy required to e...

A single electron orbits a stationary nucleus (Z=5)\left( {Z = 5} \right) .The energy required to excite the electron from third to fourth Bohr’s orbit will be-
1. 4.5 eV4.5\,eV
2. 8.53 eV8.53\,eV
3. 25 eV25\,eV
4. 16.53 eV16.53\,eV

Explanation

Solution

Atomic structure is the structure of an atom that consists of a nucleus (the centre) and protons (positively charged) and neutrons (neutral). The electrons, which are negatively charged particles, revolve around the nucleus's core.

Complete answer:
The planetary model was proposed first by the Bohr Model of the hydrogen atom, but the electrons were assumed later. The concept was that the atoms' structure could be quantified. According to Bohr, electrons orbited the nucleus in complex orbits or shells with a fixed radius. Only the shells with the radius defined by the equation below were permitted, and electrons were not permitted to exist within them. The approved atomic radius value can be determined using the following equation:
r(n)=n2×r(1)r\left( n \right) = {n^2} \times r\left( 1 \right)
Where,
The number nn is a positive integer.
The smallest radius allowed for the hydrogen atom, also known as the Bohr's radius, is r(1).r\left( 1 \right).
r(1)=0.52910−10mr\left( 1 \right) = {0.52910^{ - 10}}m is the magnitude of Bohr's radius.
By considering electrons in circular, quantized orbits, Bohr measured the energy of an electron in the nth stage of hydrogen:
E(n)=−1n2×13.6 eVE\left( n \right) = - \dfrac{1}{{{n^2}}} \times 13.6\,eV [Where, the lowest possible energy of a hydrogen electron E(1)E\left( 1 \right) is 13.6 eV13.6{\text{ }}eV .]
Now, coming to the question, the ground ionisation energy of hydrogen is : 13.6eV13.6eV
En = − 13.6 × Z2/n2En{\text{ }} = {\text{ }} - {\text{ }}13.6{\text{ }} \times {\text{ }}{Z^2}/{n^2}
E = E4− E3E{\text{ }} = {\text{ }}{E_4} - {\text{ }}{E_3}
E{\text{ }} = {\text{ }} - {\text{ }}13.6{\text{ }} \times {\text{ }}{5^2}{\text{ }}\\{ 1/{4^2}{\text{ }} - {\text{ }}1/{3^2})
E{\text{ }} = {\text{ }} - {\text{ }}13.6{\text{ }} \times {\text{ }}25{\text{ }} \times {\text{ }}\left\\{ { - 7/144} \right\\} = 16.53eV16.53eV
So, the correct option is: (4) 16.53 eV16.53\,eV .

Note:
The obtained energy is always negative, with the most negative value being for the ground state n =1n{\text{ }} = 1 . The energy of an electron in orbit is proportional to the energy of an electron totally isolated from its nucleus, n=∞n = \infty which is equal to 0 eV0{\text{ }}eV . Since an electron in a fixed orbit around the nucleus is more stable than an electron far away from the nucleus, the energy of an electron in orbit is always negative.