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If n is the orbit number of the electron in a hydrogen atom, the correct statement among the following is
A) Electron energy increases as n increases.
B) Hydrogen emits infrared rays for the electron transition from n=∞ to n=1.
C) Electron energy is zero for n=1.
D) Electron energy varies as ${n^2}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: The allowed energies of the electron in the hydrogen atom are:
${E_n} = - \dfrac{{13.6}}{{{n^2}}}eV$
Here n is called the principle quantum number. The values ${E_n}$ are the possible value for the total electron energy in the hydrogen atom. Electron energy increases with orbit number.
In these types of questions, we will check the statements whether they are in the correct relation with the formula or not.

Complete Solution:
Energy of electron is given by-
${E_n} = - \dfrac{{13.6}}{{{n^2}}}eV$
For the hydrogen atom, the energy levels of the electron only depends on the value of n.
As the value of n increases, the magnitude of energy of the electron decreases but due to the negative sign, energy of the electron increases.
Hence, Electron energy increases as n increases.

Option A is correct.

Note: It should be noted that electron energy can never be zero for n=1 as it has value equals to -13.6eV for n=1.
The energy is always going to be a negative number, and the ground state, n=1 has the most negative value. This is because the energy of an electron in orbit is relative to the energy of an electron that has been completely separated from its nucleus, n=∞ which is defined to have an energy of 0 eV
Electron energy does not vary as ${n^2}$. The energy of an electron in a Hydrogen atom is decided by principal quantum number (n) only. The number of degenerate orbitals is equal to the number of orbitals in a principal quantum level and is given by ${n^2}$.