Modern physics, imaginary particle

[Coolstorybro]

Homework Statement

The energy level scheme for the mythical one-electron element crazyidium(the names not really relevant). The potential energy of an electron is taken to be zero at an infinite distance from the nucleus (a) How much energy does it take to ionize an electron from the ground state (b) A 15 eV photon is absorbed by the crazyidium atom, what are the possible wavelengths can be emitted photons have (c) what will happen if a photon with energy of 8eV strikes a crazyidium atom? why? (d) if photons emitted from crazyidium transitions n=4 to n=2 and from n=2 to n=1 will eject photoelectrons from an unknown metal, but the transition n=3 to n=2 will not, what are the limits (maximum and minimum possible values) of the work function of the metal (e) if a 40eV photon strikes the electron in the ground state what will be the deBroglie wavelength of the ejected electron. THIS ATOM IS NOT HYDROGEN!

Homework Equations

KE= E - $\phi$
KE=(1/2)mv²
h=pλ
E=hf
M_e = 9.11x10^-31
M_p = 1.67x10^-27
E=pc
h=6.63x10^-34 J(seconds)
or = 4.14x10^-15eV(seconds)
e=1.6x10^-19C
hc=1240eV(nm)

The Attempt at a Solution

My work is a mess, and this sheet is old so i can't exactly read it

 

[DrClaude]

The energy level scheme for the mythical one-electron element crazyidium(the names not really relevant).

Can't help much if I don't see the level scheme.

Homework Equations

KE= E - $\phi$
KE=(1/2)mv²
h=pλ
E=hf
M_e = 9.11x10^-31
M_p = 1.67x10^-27
E=pc
h=6.63x10^-34 J(seconds)
or = 4.14x10^-15eV(seconds)
e=1.6x10^-19C
hc=1240eV(nm)

You seem to be missing the most important one: What is the equation for the energy levels of a hydrogenic atom?

 

[Coolstorybro]

sorry
n=infinity___________ 0eVn=4_______________-2eV
n=3_______________-5eV

n=2_______________-10eVn=4_______________-20eV

I'm not sure what the equation is for the energy levels of a hydrogenic atom,

E=E(initial)(1/(n(initail)squared) + 1/(n(final)squared))

I'm not sure about that equation, i can't remember that equation or if that's the right one or not

 

[DrClaude]

sorry
n=infinity___________ 0eV


n=4_______________-2eV
n=3_______________-5eV

n=2_______________-10eV


n=4_______________-20eV

I'm not sure what the equation is for the energy levels of a hydrogenic atom,

E=E(initial)(1/(n(initail)squared) + 1/(n(final)squared))

I'm not sure about that equation, i can't remember that equation or if that's the right one or not

Looks like you were aiming at the Rydberg formula for the energy of transitions:
$$
\Delta E = h c R \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)
$$
with $R$ the Rydberg constant. But since the energy of all levels all already given in the problem, you actually don't need that equation.

That level scheme contains all the information need to solve the problem. You'll have to be more specific as to where you have problems.

 

[paranormal]

, but i think i did it right.

I would like to clarify that there is no such thing as a "mythical one-electron element" called crazyidium. This is a made-up element for the purpose of this homework problem.

(a) In order to ionize an electron from the ground state, it would require the energy equal to the ionization energy of crazyidium. This value is not given in the problem, so it cannot be calculated.

(b) The energy of a photon is given by E=hf, where h is Planck's constant and f is the frequency of the photon. Since the energy of the photon is 15 eV, we can find the frequency using the equation E=hf. We know that E=15 eV and h=4.14x10^-15 eV*s, so we can rearrange the equation to solve for f. This gives us f=15/4.14x10^-15 = 3.62x10^15 Hz.

Now, we can use the equation c=λf to find the possible wavelengths of the emitted photons. Since c is the speed of light and we know the frequency from the previous calculation, we can rearrange the equation to solve for λ. This gives us λ=c/f = (3.00x10^8 m/s) / (3.62x10^15 Hz) = 8.29x10^-8 m = 82.9 nm.

(c) If a photon with energy of 8 eV strikes a crazyidium atom, it will not have enough energy to ionize the electron. This is because the ionization energy of an electron in the ground state is likely higher than 8 eV. Therefore, the photon will either be absorbed or it will not cause any change in the atom.

(d) The work function of a metal is the minimum amount of energy required to eject an electron from the metal's surface. In this case, the minimum work function would be the ionization energy of the electron in the n=4 state. This is because in order for the electron to be ejected, it must first be excited to the n=4 state and then have enough energy to overcome the work function. The maximum work function would be the energy difference between the n=3 and n=2 states, as this transition does not eject any photoelectrons.