CSSA Question on Ionising Energy (Quanta to Quarks) (1 Viewer)

astron

Member
Joined
Apr 3, 2003
Messages
47
Gender
Undisclosed
HSC
N/A
Does anyone actually know where Ionising Energy is supposed to be in the Quanta to Quarks course?

If so, how did you do the question?
 

Chand

Reflect the lights
Joined
Mar 29, 2003
Messages
871
Location
In the heavens
Gender
Female
HSC
2003
im not sure, i didnt to that Q or finished the topic. A suggestion is that it would somehow be related to the ionising ability in alpha, beta and gamma, and thus how they are used in medicine/agriculture/industry...hope that helps..
 

astron

Member
Joined
Apr 3, 2003
Messages
47
Gender
Undisclosed
HSC
N/A
actually, it was a calculation question, asking for the ionising energy of a hydrogen atom
 

Saul

*ss
Joined
Jul 25, 2003
Messages
340
Gender
Undisclosed
HSC
2003
you use that guys equation... ummm... forget his name, but it has rhybergs constant, and you work out the energy if it goes to an infinite quantum number level thingy.
 

jims

Member
Joined
Jul 8, 2003
Messages
127
Location
Sydney
Gender
Male
HSC
2003
its the energy that's needed to completely remove an electron from an atom.

for this question, you use that equation...yeh i forget the name too...but you put in n(final) as infinity and n(initial) as 1.
now you know the wavelength of the photon needed (which turns out to be 1/-R).
then put that into E=hc/wavelength (it comes out nicely as -hcR) and that should get you -13.6eV
 

deyveed

School Leaver
Joined
Oct 13, 2002
Messages
639
Gender
Male
HSC
2003
ohh
You know this eV stuff, do you just get the energy from E=hf and times that with 931.5 or something to change it to eV?
 

PoLaRbEaR

The Bear
Joined
Apr 13, 2003
Messages
253
Location
Sydney
Gender
Male
HSC
2003
Converting joules to eV:
number of joules divided by 1.6x10^-19
 
Last edited:

Mathematician

Member
Joined
Nov 3, 2002
Messages
188
.....

For some reason the 1st ionising energy is the electrons total energy in the stationary state given by

E(n)= - ( 2pi k^2 e^4 m)/(h^2 n^2)

I got this from the excel - it works out to be 13.6eV

but i dont know why or if we are required to know this formula along with the other long ones in the xcel book.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top