Please note that I've scaled up the ordinate value by a factor of 100 to make the difference clear.
As mentioned in the part (b) of the question, for large values of time, the graph is almost linear. This is fine. But why does it attain this linear behaviour? And why must the slope of this must be dependant only on Ag-110?
The answers I found for each part are as follows and they also match with the solution given in my book:
Half life of Ag-110 is 24.4 s, and
half life of Ag-108 is 144 s.
Is that because of the fact half life of Ag-110 is much less than that of Ag-108, at larger times, we don't have much of Ag-110 and the graph approaches the decay curve of Ag-108?
I'm having some doubts on the following problem. If possible, could you clarify them? :
> The K-beta X-ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionise an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom.
The energy needed to knock out an electron from the K shell is nothing but the ionisation energy. So simply, I arrived at the answer 16 eV but it's incorrect?
@JohnRennie: Could you explain what the author is looking for?
The energy needed to ionise an argon atom is the energy required to remove an electron from the outermost orbital, not the energy needed to remove a K electron.
And the ionisation energy is the energy required to remove an electron from the outermost shell to infinity. In this case the outermost shell is the M shell (it's actually the 3p but we'll ignore the difference between the 3s and 3p for now).
@JohnRennie I think, you're trying to say, that the energy need to kick an electron out of the K shell is same as the energy of the K-beta line plus the "minimum energy" the question supplied. Did I understand you're idea completely, sir?
