The ionisation energy of `H` atom is `21.79xx10^(-19) J`. The the value of binding energy of second excited state of `Li^(2+)` ion
A. `3^(2)xx21.7xx10^(-19)J`
B. `21.79xx10^(-19) J`
C. `1/3xx21.79xx10^(-19) J`
D. `1/3^(2)xx21.79xx10^(-19) J`
Correct Answer - A
Ratio of kinetic energy `K_(1)/K_(2)=((Z_(1)//n_(1))^(2))/((Z_(2)//n_(2))^(2))`
Since `n_(1)=n_(2)=2` & `Z_(1)=1`
for `H, Z_(2)=2` for `He^(+) implies K_(1)/K_(2)=1/4`
Correct Answer - A-q,r ; B-p,s ; C-p,s,t ; D-q
(A)B.E. of `He^+` atom =`(13.6xx2^2)/n^2` n=1,2,3…
Hence it can be 13.6 ev , 3.4 ev both
( B)In `7to3` transition `Deltan=7-3=4`...
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
`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)]`