If the binding energy of `2^(nd)` excited state of a hydrogen like sample os `24 eV` approximately, then the ionisation energy of the sample is approx

Given ,total energy of the electron `E=-4.6 EV` (i) Kinetic energy of electron K=-(Total energy E,) `rArr " " =-E=-(-4.6)=4.6 eV` (ii) de- Brogile wavelength `lambda_(d)=(h)/(sqrt(2 mK))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx4.6xx1.6xx10^(-19)))` `0.57xx10^(-9)m=0.57 nm`

Correct Answer - C The energy rquired to removed electron in the first excited state (n=2) The energy of the photon `Rhc((1)/(1^(2))-(1)/(n^(2)))` `rArr" " 13.6((1)/(4)-(1)/(4))=13.6((4-1)/(4))` `" " =13.6 xx(3)/(4)=(40.8)/(4)=10.2eV`

Correct Answer - D In ground state , kinetic energy `= 13.6 eV`, Potential energy `= -27.2 eV` In first excited state, kinetic energy `= 3.4 eV`, Potential energy `= -...

Correct Answer - B `E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For second excited state to first excited state `E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))` For first excited state...

Correct Answer - A::C::D `122.4 = (13.6Z^(2))/(1^(2)) rArr Z = 3, 91.8 = 122.4[1-(1)/(4)]` So an electron of KE `91.8 eV` can transfer its energy to this atom.

Correct Answer - A `10= .^(n)C_(2) implies n=5`, then 5 orbits are involved upon coming to second excited state so `n^(th)` excited state is `6^(th) [2^(nd), 3^(rd), 4^(th), 5^(th), 6^(th)]`

Correct Answer - B B.E.`=(21.79xx10^(-19)(3)^(2))/((3)^(2))=21.79xx10^(-19)J`

Correct Answer - B Mole of `H_(2)` present in one litre `=(PV)/(RT)=(1xx1)/(0.0821xx298)=0.0409` Thus, energy needed to break H-H bonds in `0.0409` mole `H_(2)=0.0409xx436=17.83 kJ`. Also energy needed to excite one H...

Correct Answer - `Z = 3`

Correct Answer - `n = 7 ; Z = 3`