In a sample of hydrogen atoms, all the atoms exist in two energy levels A and B. A is the ground level and B is some higher energy level. These atoms absorb photons of energy 2.7 eV and attain a higher energy level C.After this, these atoms emit photons of six different energies. Some of these photon energies are higher than 2.7 eV, some equal to 2.7 eV and some are loss than 2.7 eV.
The principal quantum number corresponding to energy level B is

Explanation

Level C must correspond to n=4 since the total number of possible transitions from level 4 to lower levels is 1+2+3+4 =6.The level corresponds either to n=2or n=3.The energies of all the photons involved in the transitions are (in eV):`E_(4),_(2)=13.6 Z^(2)((1)/(2^(2))-(1)/(4^(2)))=2.55Z^(2)`,`E_(4),_(3)=0.66Z^(2)`,`E_(4),_(1)=12.75Z^(2)`,`E_(3),_(2)=1.89Z^(2)`,`E_(3),_(1)=12.1Z^(2)`,`E_(2),_(1)=10.2Z^(2)`.The only possible choices are (a) Z=1 and `E_(4),_(2)=2.55ZeV approx 2.7eV`(b) Z=2 ,`E_(4),_(3)=2.64ZeV approx 2.7eV`.The choice Z=2 however is consistent with the fact that some products have energy less than 2.7 eV.(i) Thus, Z=1 and the quantum number of level B is n=2.The ionisation energy is ,`13.6eVxx1^(2)=13.6eV`.(iii) The maximum energy of the photons is `12.75 eV (E_(4),_(1)) while the minimum is `0.66eV(E_(4),_(3))`.