The electron of a hydrogen atom revolves around the proton in a circular $n^{t h}$ orbit of radius $r_{n}=\dfrac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}}$ with a speed $V_{n}=\dfrac{e^{2}}{2 \varepsilon_{0} n h} .$ The current due to the circulating charge is proportional to:-
(A) $e^{2}$
(B) $e^{5}$
(C) $e^{3}$
(D) $e^{6}$
Answer
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Hint: We know that electrons are the negatively charged particles of atoms. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus. Electrons are extremely small compared to all of the other parts of the atom. A proton is a subatomic particle with a positive charge. It is found in the atomic nucleus. Atoms consist of smaller particles called protons, neutrons, and electrons. An atom is the smallest unit of ordinary matter that forms a chemical element. Every solid, liquid, gas, and plasma are composed of neutral or ionized atoms. Atoms are extremely small, typically around 100 picometers across. Based on this concept we have to write the answer.
Complete step by step answer:
From given values of $\mathrm{r}_{\mathrm{n}}$ and $\mathrm{V}_{\mathrm{n}}$ we can effectively find the time period of revolution as $\mathrm{T}=\dfrac{\mathrm{r}_{\mathrm{n}}}{\mathrm{V}_{\mathrm{n}}}$ and the current can be equated as $I=\dfrac{q}{T}=\dfrac{e}{T}=\dfrac{e V_{n}}{r_{n}}$
Putting all values from the given data, we can determine that $I=\dfrac{\pi \mathrm{me}^{5}}{2 \epsilon_{0}^{2} \mathrm{n}^{3} \mathrm{h}^{3}}$
Therefore, I $\propto \mathrm{e}^{5}$
Hence, the correct answer is Option B.
Note: We should know that hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force. Atomic hydrogen constitutes about 75 percent of the baryonic mass of the universe. Hydrogen only has one electron in its lowest energy level. This is a very unstable arrangement, and hydrogen gas undergoes a variety of reactions so as to reach a stable electron configuration where its energy level is either empty of electrons, or filled with electrons. The ground state of hydrogen is the lowest allowed energy level and has zero angular momentum. However, it is the most stable state in which a single electron occupied the 1s atomic orbital.
Complete step by step answer:
From given values of $\mathrm{r}_{\mathrm{n}}$ and $\mathrm{V}_{\mathrm{n}}$ we can effectively find the time period of revolution as $\mathrm{T}=\dfrac{\mathrm{r}_{\mathrm{n}}}{\mathrm{V}_{\mathrm{n}}}$ and the current can be equated as $I=\dfrac{q}{T}=\dfrac{e}{T}=\dfrac{e V_{n}}{r_{n}}$
Putting all values from the given data, we can determine that $I=\dfrac{\pi \mathrm{me}^{5}}{2 \epsilon_{0}^{2} \mathrm{n}^{3} \mathrm{h}^{3}}$
Therefore, I $\propto \mathrm{e}^{5}$
Hence, the correct answer is Option B.
Note: We should know that hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force. Atomic hydrogen constitutes about 75 percent of the baryonic mass of the universe. Hydrogen only has one electron in its lowest energy level. This is a very unstable arrangement, and hydrogen gas undergoes a variety of reactions so as to reach a stable electron configuration where its energy level is either empty of electrons, or filled with electrons. The ground state of hydrogen is the lowest allowed energy level and has zero angular momentum. However, it is the most stable state in which a single electron occupied the 1s atomic orbital.
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