Number of visible lines when an electron returns from 5th orbit to ground state in H spectrum :

Correct Answer - B `n^(th)` excited state `=.^(n+1)C_(2) =(n(n+1))/(2) =Sigma n`

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

Correct Answer - A For n=2, using, `E_(n)=(-13.6eV)/(n^(2))E_(2)=-(13.6)/((2)^(2))=-3.4 eV` For n=1,`E_(1)=-13.6 eV` `:. " " E_(1rarr2)=-3.4-(-13.6)=10.2ev`

Correct Answer - C `vprop(Z)/(n)` since Z and n both have become twice, velocity of electron in second orbit of `H^(+)` will be v.

Correct Answer - D The ground state of hydrogen (n=1) is represented by K, the first excited state (n=2) is represented by L,the second excited state (n=3) is represent byb...

Correct Answer - D Here, `r_1=5.30xx10^(-11)m,v_1=2.2xx10^6ms^(-1)` In the second excited state, `r_n=n^2r_1,v_n=(v_1)/(n)` `therefore r_2=4r_1=4xx5.30xx10^(-11)m=2.12xx10^(-10)m` and `v_2=(v_1)/(2)=(2.2xx10^(6))/(2)ms^(-1)=1.1xx10^(6)ms^(-1)`

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 `V=2.188xx10^(6) Z/n m//s` Now, `V prop Z/n` so, `V_(Li^(2+))/V_(H)=-(Z_(1)//n_(1))/(Z_(2)//n_(2))=(3//3)/(1//1)=1` or, `V_(Li^(2+))=V_(H)`

Correct Answer - D Cation at `3^(@) gt 2^(@) gt 1^(@)`

Correct Answer - `94.7 nm, 96.9 nm, 102.2 nm, 121 nm`