An excited `.^(69)Zoverset(*)n` Nucleus at rest decays to ground state `.^(69)Zn` by way of emitting a gamma photon. The energy equivalent of differen

To determine the Q value for this decay, we first need to find the mass of the product nucleus `.^(12)C` in its excited state. In the ground state, `.^(12)C` has...

Correct Answer - C Energy difference in hydrogen atom `= 13.6 - (13.6)/(n^(2)) = 12.09 rArr n^(2) ~~ 9 rArr n = 3`

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 - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T)...

Correct Answer - isotope

Correct Answer - (a) `Z_(A)-Z_(B)=4;A_(A)-A_(B)=4` (b) `(9N_(0))/(2)` (c) 6 hr

Correct Answer - 227.62u

Correct Answer - 1.008968u

Correct Answer - D `(d)` We have `betaalpha+gammaalpha+alphabeta=0` `Delta=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma,gammaalpha,alphabeta),(gammaalpha,alphabeta,betagamma),(alphabeta,betagamma,gammaalpha):}|` `=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma+gammaalpha+alphabeta,gammaalpha,alphabeta),(gammaalpha+alphabeta+betagamma,alphabeta,betagamma),(alphabeta+betagamma+gammaalpha,betagamma,gammaalpha):}|` [using `C_(1)toC_(1)+C_(2)+C_(3)`] `=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(0,gammaalpha,alphabeta),(0,alphabeta,betagamma),(0,betagamma,gammaalpha):}|=0` [all zero property].