How Do Wavelength Changes Affect Electron Energy Liberation?

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The discussion centers on a physics problem involving the liberation of electrons from a metal when exposed to light of varying wavelengths. When a light beam strikes the metal, it liberates 1 eV electrons, but when the wavelength is halved, it liberates 5 eV electrons. The key formula referenced is E = hc/λ, indicating that halving the wavelength doubles the energy of the photons. The concept of work function is also introduced, which is the minimum energy required to release an electron. The participants express confusion due to the lack of a textbook and seek clarification on the relationship between photon energy and electron liberation.
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Homework Statement


When a certain light beam strikes a certain metal, 1 eV electrons are liberated. If the beam is replaced by one of half the original wavelength, 5 eV electrons are liberated. What is the energy of the photons in the original beam?


Homework Equations


Part of my confusion with this problem is I don't know of any formula that can tell me this information. The only formula that references wavelength that I've seen in class is E = hc/L where hc is a conversion factor equal to 1240 eV*nm.

The Attempt at a Solution


I guess I need to find out the relationship between liberating electrons and energy. When the wavelength is reduced to 1/2 L, it increases the amount of liberated electrons to 5 eV. Unfortunately, my class has no textbook, so it's very hard for me to figure this out on my own. If anyone can help me out, I'd very much appreciate it!
 
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sdecosta100 said:

Homework Statement


When a certain light beam strikes a certain metal, 1 eV electrons are liberated. If the beam is replaced by one of half the original wavelength, 5 eV electrons are liberated. What is the energy of the photons in the original beam?


Homework Equations


Part of my confusion with this problem is I don't know of any formula that can tell me this information. The only formula that references wavelength that I've seen in class is E = hc/L where hc is a conversion factor equal to 1240 eV*nm.

The Attempt at a Solution


I guess I need to find out the relationship between liberating electrons and energy. When the wavelength is reduced to 1/2 L, it increases the amount of liberated electrons to 5 eV. Unfortunately, my class has no textbook, so it's very hard for me to figure this out on my own. If anyone can help me out, I'd very much appreciate it!

I hope you know things like the work function of a metal - that minimum amount of energy an incoming photon [or electron] must give an atom before an electron can be released.
Also your expression E = hc/λ shows that if you halve the wavelength [λ] you double the energy.

These problems can be compared to a money problem.

You arrive at an establishment with $X. After paying the entry fee [analogy to Work Function] you get to go in, with $1 still in your pocket.
A friend arrives with twice as much money as you had [$2X]. After paying the entry fee they get to go in, with $5 still in their pocket.
How much money did you approach with?
You can also work out how much money they had, and what the entry fee was.
 
sdecosta100 said:

Homework Statement


When a certain light beam strikes a certain metal, 1 eV electrons are liberated. If the beam is replaced by one of half the original wavelength, 5 eV electrons are liberated. What is the energy of the photons in the original beam?


Homework Equations


Part of my confusion with this problem is I don't know of any formula that can tell me this information. The only formula that references wavelength that I've seen in class is E = hc/L where hc is a conversion factor equal to 1240 eV*nm.

The Attempt at a Solution


I guess I need to find out the relationship between liberating electrons and energy. When the wavelength is reduced to 1/2 L, it increases the amount of liberated electrons to 5 eV. Unfortunately, my class has no textbook, so it's very hard for me to figure this out on my own. If anyone can help me out, I'd very much appreciate it!

The expression you might be after is Emax = hf - W or Emax = h(f - fo)
 
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