Lets say we give an electron in ground state in an $\rm H$-atom 11eV. What will happen?
My intuition says that the electron will take the amount of energy to reach the next orbital and the rest is 'ignored'. Hence it reaches the 2nd orbital in this case. But my teacher said that the entire energy is 'ignored' ie the electron doesn't jump. So it stays in ground state.
Can anyone clear this conundrum?
Transitions can only occur if the energy imparted to the atom is a possible energy difference between energy levels. This is why for instance the spectrum of hydrogen is discrete rather than continuous.
In your case you have some pulse or other source of energy which arrives with 11ev but there are no integers $n_1,n_2$ so that $E_{n_1}-E_{n_2}=11$eV. You can verify this easily since you cannot find a pair of integers $n_1,n_2$ so that $$ E_{n_1}-E_{n_2}=-13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) = 11\text{eV}. $$ Thus, in your example, no transition would happen: the pulse would not be absorbed and would go through the atom with no effect. Turning this around you will never see an emission line with a wavelength corresponding to 11eVs in a hydrogen spectrum.
In practice a pulse would not have exactly $11$eV in energy: there would be an energy distribution centred around 11eV, with some probability of having a little more or a little less so it is theoretically possible for a pulse to induce transitions if the energy of the pulse (by which we mean the average energy) is not exactly the energy difference between two levels. However, this depends on the details of how you input energy in the system.
Also the energy levels are not exactly sharp: they actually have a energy in a very narrow range around the recognize value as a result of various broadening effects.
