Solving Atom Ionization with a 50 nm Wavelength

In summary, a hydrogen atom with an electron in the fundamental state can be ionized by a radiation with a wavelength of 50 nm. The excess kinetic energy of the electron can be calculated by subtracting the ionization energy of 13.6 eV from the energy of the photon, which is 24.8 eV. This results in an excess kinetic energy of 11.24 eV, which is enough to ionize the atom. The excess kinetic energy can also be calculated in joules by using the formula E = hc/λ, which yields 1.78x10^-18 J.
  • #1
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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. ?
 
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  • #2
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...
 
  • #3
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.
 
  • #4
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
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?
 
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  • #6
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
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
Yes E = h(c/lambda) or hf.
Momentum = h/lambda or hf/c
 
  • #9
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
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
o ok that helps a lot and saves a lot of time thanks :)
 

Related to Solving Atom Ionization with a 50 nm Wavelength

1. What is atom ionization and why is it important?

Atom ionization is the process of removing one or more electrons from an atom, resulting in a positively charged ion. It is important because it allows scientists to manipulate and control the behavior of atoms, which is crucial in fields such as nanotechnology, materials science, and quantum computing.

2. How does a 50 nm wavelength help in solving atom ionization?

A 50 nm wavelength falls within the ultraviolet range, which is known to have a strong ionizing effect on atoms. This means that when a beam of light with a 50 nm wavelength is directed at an atom, it has a high probability of removing one or more electrons from that atom, leading to ionization.

3. Can a 50 nm wavelength be used to ionize all types of atoms?

No, the effectiveness of a 50 nm wavelength in ionizing atoms depends on the specific electronic structure of the atom. Atoms with larger atomic radii and lower ionization energies tend to be more easily ionized by a 50 nm wavelength compared to smaller atoms with higher ionization energies.

4. What are the potential applications of using a 50 nm wavelength to solve atom ionization?

The ability to precisely control and manipulate the ionization of atoms using a 50 nm wavelength has a wide range of potential applications. It can be used in the development of new materials, studying the properties of individual atoms, and in the creation of advanced technologies such as quantum computers.

5. Are there any limitations or challenges when using a 50 nm wavelength to solve atom ionization?

One limitation is that the ionization process can also affect other nearby atoms, leading to unintended consequences. Additionally, the use of a 50 nm wavelength requires highly specialized equipment and complex techniques, making it challenging to implement in certain research settings.

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