Solving Atom Ionization with a 50 nm Wavelength

[stickplot]

Homework Statement

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?
Round up your answer to the nearest hundredth.

Homework Equations

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

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. ?

 

[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 ?

think about it...

 

[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 doesn't 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 I am not sure if that's for kinetic energy.

 

[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...

 

[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?

 

[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".

 

[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?

 

[Delphi51]

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

 

[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?

 

[Redbelly98]

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

E_photon = 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.

 

[stickplot]

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