However, the boundary of 2s has a spherical surface around the nucleus and another spherical shell around the nucleus. Thus there is a node between the two spherical surfaces describing 2s surface while just one spherical surface describes 1s orbital.
Q3. The probability distribution function is obtained by multiplying the square of the wave function with 4r2. Even though the radial part of the wave function is maximum at r = 0; the value of the probability distribution function will be zero only as r = 0. Thus the maximum probability of finding an electron can be at some other value of r; even though the wave function has its maximum value at r = 0. Thus it can be seen that there is no contradiction in probability of finding an electron being maximum at r = 0.529 A and value of the corresponding wave function being maximum at r = 0. Hence the two statements are compatible.
Q4. This is because alkali metals are multi electron system unlike hydrogen, which is one electron system. Therefore, in case of hydrogen only electron – nucleus interaction is there and the atomic spectrum of hydrogen is due to electronic transition of just one electron across different excitation states. In case of alkali metals, there being many electrons, there exists interaction among electrons as well besides the electron – nucleus interaction. Further there are many more excitation states and electronic transitions in case of alkali metals (because of many electrons being there) than in case of hydrogen atom. Therefore, the atomic spectrum of an alkali metal is much more complex than that of hydrogen.
Q7. First ionization energy of an atom is the value of energy required to pull out the outermost electron from the atom. Its value increases as one traverse from left to right in a period of the periodic table. This is because as one moves from left to right in a period the number of protons I the