# Quantum physics

1. Feb 15, 2009

### stickplot

1. The problem statement, all variables and given/known data

A hydrogen atom has an electron in the fundamental state.
a. Show that a radiation with λ = 50 nm will ionize the atom.
b. What will be the excess kinetic energy of the electron in joules?

2. Relevant equations
1/lambda= R(1/n^2-1/k^2) (im not sure if this is the equation to be used)

3. The attempt at a solution

1/5x10^-8= 1.097x10^7(1/1^2-1/n^2)
2x10^7=1.097x10^7(1/1^2-1/n^2)
1.823154057=1-1/n^2
.8231540565=1/n^2
1.214839424=n^2
1.1= n
how would this show that atom ionizes?
b. ?

2. Feb 15, 2009

### sanjibghosh

If the value of 'R' is correct ,then 'n=1' because n=always integer ,so after that I think you can able to calculate it what's ans. of part 'b'...
see, ionized atom means 'neutral atom minus an electron' ,and this will happen when the electron has +ve energy ............

........n=1,2,3.....are the possible energy levels .........there are no levels in-between them........so if you have got n=1.1 ........is it possible that the electron is bound ?

3. Feb 16, 2009

### stickplot

ooo ok. i understand the question now, but would 1.1 just round off to 1? and if it would how is that ionized, if the electron has -eV? and if it doesnt round off then what would happen because i know only integers are for n.
and for b? because i know to find the energy of the orbit is -13.6 eV/n^2 but its asking for kinetic energy, and im not sure if thats for kinetic energy.

4. Feb 16, 2009

### sanjibghosh

Total energy=neu*h,where 'neu'=c/lamda...after that if you just subtract 13.6 energy (if fundamental state means n=1) then you will get the kinetic energy of the electron ..... but why? Don't look at the given relation....U have a question ,ans.. it..
All the statements may be or may not be correct........

5. Feb 16, 2009

### stickplot

ok.
so what if its in a bound state,
doesnt it need to jump from a bound state to a free state to be an integer because it needs to have a positive value.
and i used the formula that you showed me, but what is h?
i just did neu=c/lambda-13.6 and i got 1.1 but that is not the kinetic energy cus all "neu" is, is the frequency right so then what is that?

Last edited: Feb 16, 2009
6. Feb 16, 2009

### Delphi51

Another way of looking at this is to convert the -13.6 eV of the n=1 state into Joules. The ionization energy is +13.6 eV, enough to bring it up to zero or beyond.
Also find the energy of the photon in Joules. This is greater than the ionization energy, so part 1 is done. Just subtract to get the "excess energy".

7. Feb 16, 2009

### stickplot

ok. i got that. but how do i get the energy of the photon?
is it h(c/lambda)?
c= speed of light
lambda= wavelength?

8. Feb 16, 2009

### Delphi51

Yes E = h(c/lambda) or hf.
Momentum = h/lambda or hf/c

9. Feb 16, 2009

### stickplot

ok i think i got it. how does this look
4.14x10^-15(c/5x10^-8)= 24.84 eV
24.48-13.6= 11.24 eV of excess kinetic energy
and it would ionize because it is now positive.
and for B would i just do the same thing but in joules?
6.6x10^-34(c/5x10^-8)= 3.96^-18
3.96^-18-2.178961169x10^-18= 1.78x10^-18 J of excess kinetic energy?

10. Feb 17, 2009

### Redbelly98

Staff Emeritus
Looks good. Alternatively, one could convert the 11.24 eV from (A) into Joules.

By the way, a useful number to keep handy in your notes (or even memorize) is
h c = 1240 eV nm​
So for example
Ephoton = h c / λ
= (1240 eV nm) / (50 nm)
= (1240/50) eV ( Cancelled the nm/nm units )
= 24.8 eV
It saves on having to carry around a lot of exponential factors like 10-15 and so on.

11. Feb 17, 2009

### stickplot

o ok that helps a lot and saves a lot of time thanks :)