A deuterium nucleus at rest absorbs a Gamma ray photon of energy 2.21 MeV and splits into a proton and a neutron. The neutron was found to be at rest

Correct Answer - A `("Energy loss")/("Initial KE") = (4m_(1)m_(2))/((m_(1) + m_(2))^(2)) = (4(1)(2))/((1+2)^(2)) = 8/9` After `1^(st)` collision `DeltaE_(1) = 8/9 E_(0)` After `2^(nd)` collision `DeltaE_(2) = (8)/(9)E_(1)`, After `n^(th)` collision...

Correct Answer - A, C More the nucleons, more the more the binding energy & more reduced is the mas. So, `M_(2) lt 20(m_(p)+m_(n))` (Because some mass got reduced on release...

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 - 1.98MeV.

Correct Answer - 1.5eV

Correct Answer - 0.7820 MeV

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].

Correct Answer - [10 us by both]

Correct Answer - [(i) 200.68 MeV` (ii) `22.86 MWh`]

Correct Answer - `[11.2eV]`