Find the quantum of the excited state of electrons in `He^(+)` ion which on transition to first excited state emit photons of wavelengths `108.5 nm`.

Correct Answer - B `{:(s = 1,t = 0,u = - (1)/(2)","(1)/(2),),(s = 2,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 3,u = - 2"," -1"," 0","+1","+2 ,),(s = 3,t = 0,u...

Correct Answer - A (A)For `Li^(2+)`, n=6 to n=3 For H, the similar transition is 2 to 1 For `He^+` , the similar transition is 4 to 2 Energy of `4^(th)`...

Correct Answer - B S=1 t=0 `u=-1/2,1/2` No of subshell=1 , No of orbital=2 No of electrons =8 S=2 t=0,3 `u=-1/2,1/2`(for t=0) u=-2,-1,0,+1,+2(for t=3) No of subshell=2, No of orbital=7 No...

Correct Answer - x=8 We have `DeltaE=3/4xx0.85eV` as energy=0.6375 the photon will belong to brackett series (as for brackett`0.31leEle0.85`) `0.85xx(1-1/4)=13.6(1/4^2-1/n^2)` `0.85(1-1/4)=13.6/16[1/(4/n)^2] :. 4/n=1/2 implies n=8` Hence x=8

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 Infrared lines =total lines -visible lines -UV lines `=(6(6-1))/(2)-4-5 =15-9=6` `("visible lines" =4 6 rarr2, 5 rarr2, 4 rarr 2, 3 rarr 2)` " " `("UV lines"...

Correct Answer - 4 `6=(n(n-1))/(2)" "n=4` Separation energy, `E=+13.6 xx(z^(2))/(n^(2))=+13.6xx((2)^(2))/((4)^(2))=3.4 eV` l.E. of Hydrogen atom `=x xxE=13.6 eV` `x=(13.6)/(E)=4`

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

Correct Answer - (b) The two are same