The limiting line in Balmer series will have a frequency of
Atomic Structure MCQs for NEET — Chemistry Questions with Answers
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The wavelength of a spectral line for electron transition is inversely related to
The wavelength of a spectral line is inversely proportional to the energy difference between the initial and final energy levels involved in the electron transition. It is independent of the nuclear charge, electron velocity, and the number of electrons undergoing the transition.
The phenomenon of splitting of spectral lines under the influence of the electric field is known as
The Stark effect is the splitting or shifting of spectral lines of atoms or molecules due to the presence of an external electric field. It is named after Johannes Stark, who discovered this phenomenon in 1913.
If the energy of electron in H atom is given by expression, $ {- 1312 \over n^2 } kJ mol ^ {-1} $ , then the energy n2 required to excite the electron from ground state to second orbit is
In the atomic spectrum of hydrogen, the spectral lines pertaining to electronic transition of n =4 to n =2 refers to
In the atomic spectrum of hydrogen, the Balmer series corresponds to the electronic transitions from higher energy levels (n ≥ 3) to the second energy level (n=2). Therefore, the spectral lines pertaining to the transition from n=4 to n=2 belong to the Balmer series.
If ionising energy of H atom is 13.6 eV, then the second ionising energy of He should be
The second ionization energy of an element is the energy required to remove an electron from the singly ionized atom. In the case of helium, the first ionization removes one electron, leaving He+ (singly ionized helium). The second ionization energy removes the remaining electron, giving He2+. Since helium has two electrons, the second ionization energy is simply twice the first ionization energy of hydrogen, which is 13.6 eV. Therefore, the second ionization energy of helium is 54.4 eV.
The first line in the Balmer series in the H atom will have the frequency
How many electrons in $ {19}_ K $ have n =3; l = 0 ?
The maximum number of electrons in s, p and d-subshells are
The maximum number of electrons that can be accommodated in s, p, and d subshells are 2, 6, and 10, respectively. This is based on the Aufbau principle and the Pauli exclusion principle.
The number of electrons that can be accommodated in $d_{xy} $ orbital is
The $d_{xy}$ orbital can accommodate a maximum of 2 electrons with opposite spins, following the Pauli exclusion principle. This is true for all the five d-orbitals ($d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, and $d_{z^2}$).
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