An electron of a hydrogen like atom is in excited, state. If total energy of the electron is -4.6 eV, then evaluate (i) the kinetic energy and (ii) th

Correct Answer - B Kinetic energy of electron in ground state =-(Totoal energy in this state) `" " --(13.6 )eV=13.6 eV`

Correct Answer - A Here, `E=-13.6eV=-13.6xx1.6xx10^(-19)=-2.2xx10^(-18)J` `E=(-e^2)/(8piepsilon_0r)` `therefore` As orbital radius, `r=(-e^2)/(8piepsilon_0E)=(9xx10^(9)xx(1.6xx10^(-19))^2)/(2xx(2.2xx10^(-18)))=5.3xx10^(-11)m`.

Correct Answer - B Energy , `E_n=-(13.6)/(n^2)eV` In ground state energy `E_1=-(13-6)/(1^2)=-13.6eV` In first excited state energy `E_2=-(13.6)/(2^2)=-3.4eV` Then the required energy `=E_2-E_1=-3.4-(-13.6)=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 - 2 Kinetic energy of electron `= 13.6 [1-1/9] eV = 12.1 eV`.

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 `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 - C Excitation upto `n=3` is required so that visible length is emitted upon de-excitation. So required energy `=13.6 (1-1/9)=12.1 eV`

Correct Answer - D `(13.6(Z)^(2))/((3)^(2))=24` I.E. `=13.6(Z)^(2)=(24xx9)=216 ev`

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...