Pragraph : A hydrogen octum in third excited state makes a transtion to first excited state and emit
photon. This emitted photon is absorbed by He+ ion which was already in seventh excited state. After absorption
of photon He+ ion jumps from seventh excited state to heigher excited state having quantum number ns.
1. The quantum number ns of the state finally populated in He+ ion.
a) 8 b) 10 c) 16 d) 20
2. Now He+ ion jumps to lower state by imitting single photon 'P' of visible light. The energy satatis photon is as possible as close to energy of the obsorbed photon by He+ ion (as mentioned in paragraph). Then wave length of the photon 'P' is nearly.
a) 6760 Angstrom b) 5000A c) 4480A d) 3500 A
3.The ratio of K.E. of the n = 4 electron for H atom to that He+ ion is P1 and the ratio of the total energy of n = 6 electron for H atom to that of He+ ion is P2 then ratio of P1/P2 is.
a) 1 b) 9/4 c) 4/9 d) 4
29and In Qn 1
Energy of electron in 3rd exited state = -0.85eV
Enegy of electron in 1st exited state = -3.4eV
so energy released = 2.55eV
Wavelength corresponding to this enegy = 5000A so Ans b for 2
since energy of He+ ion in 7th exited state = -0.85eV
so by absorbing 2.55eV the electron in the ion will leave the atom and may go to infinity..so most probable answer for 1 i think shud be D
