The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^2 km.s-1.
c) What is the wavelength of the radiation that caused photo ejection of the electron?
For c, why can't we use the frequency of the radiation stated in b to find the wavelength of the radiation?
Textbook 1B.15
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Re: Textbook 1B.15
For part C you must use the equation
E=Work function + KE
B's answer gives us the work function and the velocity is known from part A and the mass of an electron is known. You can plug all of this in to find the energy of the photon.
Then using the equation lamda=(hc)/E we can plug in the energy we just found to find the wavelength for part c.
We cant just use the frequency of the radiation from B because we must first find the energy the photon carries and then use that energy to find the wavelength.
E=Work function + KE
B's answer gives us the work function and the velocity is known from part A and the mass of an electron is known. You can plug all of this in to find the energy of the photon.
Then using the equation lamda=(hc)/E we can plug in the energy we just found to find the wavelength for part c.
We cant just use the frequency of the radiation from B because we must first find the energy the photon carries and then use that energy to find the wavelength.
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Re: Textbook 1B.15
The wavelength found in part b relates to the required energy to eject the electron from the metal surface, while part c asks for the wavelength of the radiation (light/photons).
The equations needed to solve part c are as followed:
E(ke) = E(photon) - E(energy to remove e-)
E(ke) = 1/2 mv^2
E(photon) = hc/lambda(photon)
From part b, we've already calculated the E(energy to remove e-). We can then calculate E(ke) using the equation above and the given velocity in the question. From there, we can calculate E(photon) and the wavelength of the photon, or radiation that caused the photoejection of the electron. I hope this was helpful!
The equations needed to solve part c are as followed:
E(ke) = E(photon) - E(energy to remove e-)
E(ke) = 1/2 mv^2
E(photon) = hc/lambda(photon)
From part b, we've already calculated the E(energy to remove e-). We can then calculate E(ke) using the equation above and the given velocity in the question. From there, we can calculate E(photon) and the wavelength of the photon, or radiation that caused the photoejection of the electron. I hope this was helpful!
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