1) A moving hydrogen atomin ground state, causing radiation of several wavelengths, the longest of which is 1880 nm. From the consideration of the loss in KE in collision, what is the velocity above which the atom has to move initially??
2)A radioactive sample simultaneosly decays via two alternative processes, the mean life of one being equal to the half-life of the other (each equal to T). The time at which thepopulation of the undecayed nuclei equals sum of daughter nuclei is
(a) \left(\frac{1}{1+ln2} \right)T
(b) \left(1+ln2 \right)T
(c) \left(\frac{ln2}{1+ln2} \right)T
(d) \left(\frac{1+ln2}{ln2} \right)T
3)
113) If the propbabiity that a radioactive nuclide is undecayed after two years is p, then it's disintegration constant is (in year-1)
(a) ln\left(\frac{1}{\sqrt{p}} \right)
(b) ln\left(\frac{1}{p} \right)
(c) ln\left(\sqrt{p} \right)
(d) ln\left(p \right)
1in first q first line make it clear.....
i ve to tk dat a hydrogen atom is movin to ground state????????
11OK reposting first question:
1) A moving hydrogen atom in the ground state collides with a stationary hydrogen atom in the ground state, causing radiation of several wavelengths, the longest os which is 1880 nm. From the consideration of the loss in KE in collision, what is the velocity above which the atom has to move initially??
1srry ani got it..
u ve not left space between atom and in. dats d prob....
now i got it.......
11tapan, if u can explain ur answer, it wud be easier for everyone else
Also, i don't have the solutions.....so pls explain ur answer
623) If the propbabiity that a radioactive nuclide is undecayed after two years is p, then it's disintegration constant is (in year-1)
let λ is the constant
then no of those decaying in 2 years
N = N0e-2λ
N / N0 = e-2λ
probab of surviving 2 years = p = N/N0 = e-2λ
so -1/2 ln(P) = λ
λ = ln(1/√P)
1he has given it as loss of kinetic energy... ( due to collision)
11sorry, sister.....i've posted the complete question in post #5
622)A radioactive sample simultaneosly decays via two alternative processes, the mean life of one being equal to the half-life of the other (each equal to T). The time at which thepopulation of the undecayed nuclei equals sum of daughter nuclei is
dA/dt = -k1 - k2
A=A0 ek1+k2
now we have that
.693/k1 = 1/2 x 0.693/k2
2 k2 = k1
now can you solve it..
