Volcano blasts are one of the most powerful calamities on earth. When they blast they release large amount of mass in the air just like a jet does. Does such a blast have any effect on the course of the Earth? Does it change the Earth's orbit in any way? If not then how big a blast is needed to do it? Would a full eruption of the Yellowstone Supervolcano do it?


Question: What is the difference between the KLMN and SPDF methods of finding electronic configuration?
Answer: The KLMN(OP) method is based on electron shells, with the labels KLMN(OP) being derived from an experiment in which the spectroscopist wanted to leave room for lower energy transitions in case their were any.

K denotes the first shell (or energy level), L the second shell, M, the third shell, and so on. In other words, the KLMN(OP) notation only indicates the number of electrons an atom has with each principle quantum number ($n$).

The SPDF notation subdivides each shell into its subshells. For further information about the other quantum numbers, especially $l$, which defines the subshell, see the accepted answer to this question: What do the quantum numbers actually signify?


  The K shell can hold two electrons: $n=1,\ l=0$


When $l=0$, we have an $s$ subshell, which has one orbital $m_l=0$, with room for two electrons.


  The L shell can hold 8 elections: $n=2,\ l=0,1$


When $l=1$, we have a $p$ subshell, which has three orbitals $m_l=-1,0,+1$, with room for 6 electrons. The L shell also has an $s$ subshell. 


  The M shell can hold 18 electrons $n=2,\ l=0,1,2$


When $l=2$, we have a $d$ subshell, which has 5 orbitals $m_l=-2,-1,0,+1,+2$, with room for 10 electrons. The M shell also has $s$ and $p$ subshells.


  The N shell can hold 32 electrons! $n=3,\ l=0,1,2,3$


When $l=3$, we have an $f$ subshell, which has 7 orbitals $m_l=-3,-2,-1,0,+1,+2,+3$, with room for 14 electrons. The N shell also has $s$, $p$, and $d$ subshells.

This [table here] summarizes the relationship between the two notations for all elements. Here is an example for scandium $\ce{Sc}$:

Scandium has 23 electrons $Z=23$. Its electron configuration in KLMN notation is

K  L  M  N
2  8  9  2


Its SPDF notation (based on the aufbau principle) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2$