It is said attraction leads to loss of energy.

 

suppose an electon in 2s orbital and another in 3p orbital.

 

Doesn't the 2s electron have more energy as it is difficult to remove it from the atom as compared to 3p electron.

Attraction results in loss of energy when the attraction causes motion in a conservative field. When you hold two charges, there is attraction between them but you bind them from motion and reaching a state of less potential energy.

2s electron has less energy and hence more energy is needed to be supplied to it to ionize it then 3p. Bohr's model provides a good analysis of one electron atom energy levels. You can check that energy of a bound system(total) happens to be negative.

@ rktpro

 

How can you say that the 2s electron has less energy, when it is tightly bound to the nucleus.....

 

As 2s electron is tightly bound, it seems to me that it has more energy. So, more energy is required to tear it off the nucleus as compared to 3p electron....

I say it because I know about potential energy of a system, electrostatic potential in this case due to proton-electron pair and kinetic energy of a revolving electron. And I know that total energy is sum of kinetic and potential energy.

Here, http://dev.physicslab.org/Document.aspx?doctype=3&filename=AtomicNuclear_BohrModelDerivation.xml

@ rktpro

 

i went through the article...i think the negative sign is because of convention....

 

i think that a 2p electron has both KE and PE greater than that of a 3s electron....

Did you read what the author say about negative sign?