Physics · Atoms · NEET
The negative sign means the electron is bound to the nucleus. We take the energy of a free electron (very far away, n = infinity) as zero. To pull the electron out of the atom we must GIVE energy, so the electron inside has energy LESS than zero, that is, negative. A more negative value means the electron is more tightly bound and needs more energy to escape.
A bigger n means MORE energy (energy increases toward zero). Compare: E1 = -13.6 eV, E2 = -3.4 eV, E3 = -1.51 eV. Since -3.4 is greater than -13.6, the n=2 orbit has more energy than n=1. Students get confused because the number 3.4 is smaller than 13.6, but with the minus sign the value is actually larger. Higher orbits are less tightly bound.
Both, but with opposite signs. The energy of the electron in the ground state (n=1) is -13.6 eV. The ionisation energy is the energy needed to remove that electron from n=1 to n = infinity, which is 0 - (-13.6) = +13.6 eV. So -13.6 eV is the electron's energy and +13.6 eV is the ionisation energy. Same magnitude, opposite sign.
Use n^2 = 13.6 / |E|, taking the magnitude of the given energy in eV. Example: if E = -3.4 eV, then n^2 = 13.6/3.4 = 4, so n = 2 (first excited state). If E = -1.51 eV, then n^2 = 13.6/1.51 = 9, so n = 3 (second excited state). Always use the positive value of E inside the division.
n=1 is the GROUND state (lowest energy, -13.6 eV). The FIRST excited state is n=2 (energy -3.4 eV), and the SECOND excited state is n=3 (energy -1.51 eV). So 'first excited state' is n=2, not n=1. Many NEET numericals hide the value of n behind these words, so read carefully: excited state number = n minus 1.
Let T1 and T2 be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to Bohr's model of an atom, the ratio T1 : T2 is:
The ground state energy of hydrogen atom is -13.6 eV. The energy needed to ionize hydrogen atom from its second excited state will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For hydrogen, En = -13.6/n^2 eV, where n is the orbit number (1, 2, 3, ...). For a hydrogen-like ion of atomic number Z, it becomes En = -13.6 Z^2/n^2 eV.
In the ground state n=1, so E1 = -13.6/1^2 = -13.6 eV. This is the lowest (most negative) energy level of the hydrogen atom.
As n increases the electron sits in a larger orbit and is less tightly bound to the nucleus. Its energy rises toward zero: -13.6 eV, -3.4 eV, -1.51 eV, and so on, approaching 0 eV as n goes to infinity.
At n = infinity, En = -13.6/infinity = 0 eV. This is a free electron completely removed from the atom, which is why we call n = infinity the ionisation limit.
Yes, but you must include Z. For He+ (Z=2), En = -13.6 Z^2/n^2 = -13.6 x 4/n^2 eV. The ground state of He+ is therefore -54.4 eV, four times deeper than hydrogen.
