For hydrogen atom, the orbital/s with lowest energy is/are:
The energy of orbitals in an atom depends on the principal quantum number (n) and the angular momentum quantum number (l). For hydrogen atoms, the energy of orbitals is primarily determined by the value of the principal quantum number (n) alone, as all orbitals of the same n value have the same energy. The only exception is for multi-electron atoms, where orbital shapes and electron-electron repulsion come into play, but this is not the case for hydrogen.
The energy of an orbital in a hydrogen atom is given by the formula:
\[ E = - \frac{13.6 \, \text{eV}}{n^2} \] where \(n\) is the principal quantum number (1 for the K-shell, 2 for the L-shell, etc.).
In hydrogen, the orbitals with lower values of \(n\) will have lower (more negative) energy. Therefore, the energy of orbitals decreases as \(n\) decreases.
The 4s orbital belongs to the n=4 shell. It has a relatively higher energy compared to orbitals with a lower value of n. Hence, this orbital does not have the lowest energy in the hydrogen atom.
The 3p orbital belongs to the n=3 shell. This orbital is lower in energy than the 4s orbital, as \(n=3\) is less than \(n=4\). In hydrogen, all 3p orbitals (including 3px, 3py, 3pz) have the same energy, so the 3px orbital has lower energy than 4s. Therefore, it is a valid candidate for the lowest energy orbital.
The 3dx²−y² orbital belongs to the n=3 shell. While 3d orbitals generally have a higher energy than 3p orbitals, they still belong to the same n=3 shell. Therefore, this orbital also has lower energy than the 4s orbital, making it a valid candidate for the lowest energy orbital in hydrogen.
Similar to 3dx²−y², the 3dz² orbital also belongs to the n=3 shell. It has the same energy as the other 3d orbitals and is lower in energy than the 4s orbital. Hence, this orbital is another valid candidate for the lowest energy orbital in hydrogen.
The 4pz orbital belongs to the n=4 shell, which has higher energy than orbitals from the n=3 shell. Thus, this orbital does not have the lowest energy in hydrogen.
Based on the analysis, the orbitals with the lowest energy for a hydrogen atom are:
Therefore, the correct answer is (B), (C), and (D) only.
An amount of ice of mass \( 10^{-3} \) kg and temperature \( -10^\circ C \) is transformed to vapor of temperature \( 110^\circ C \) by applying heat. The total amount of work required for this conversion is,
(Take, specific heat of ice = 2100 J kg\(^{-1}\) K\(^{-1}\),
specific heat of water = 4180 J kg\(^{-1}\) K\(^{-1}\),
specific heat of steam = 1920 J kg\(^{-1}\) K\(^{-1}\),
Latent heat of ice = \( 3.35 \times 10^5 \) J kg\(^{-1}\),
Latent heat of steam = \( 2.25 \times 10^6 \) J kg\(^{-1}\))
Match List - I with List - II.

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
Given three identical bags each containing 10 balls, whose colours are as follows:
| Bag I | 3 Red | 2 Blue | 5 Green |
| Bag II | 4 Red | 3 Blue | 3 Green |
| Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is:
If \( \theta \in \left[ -\frac{7\pi}{6}, \frac{4\pi}{3} \right] \), then the number of solutions of \[ \sqrt{3} \csc^2 \theta - 2(\sqrt{3} - 1)\csc \theta - 4 = 0 \] is equal to ______.