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### 1B. 15C

Posted: Thu Oct 17, 2019 12:50 am
The velocity of an electron that is emitted from a metallic surface by a photon is 3.63 x 10^3 km/s.
(a) What is the wavelength of the ejected electron?
(b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 x 10^16 Hz. How much energy is
required to remove the electron from the metal surface?
(c) What is the wavelength of the radiation that caused photoejection of the electron?

For part a, I got 2.0 x 10^-10 m, and for part b I got 1.66 x 10^-17 J. In the solutions manual, the energy required to remove the electron is added to the electron's kinetic energy, and I don't understand why. Also, why can't I use the equation of lambda = h/mv?

### Re: 1B. 15C  [ENDORSED]

Posted: Thu Oct 17, 2019 10:13 am
The energy required to remove the electron is added to the electron's kinetic energy because you need the total energy that caused the electron not only to be ejected but to be ejected at 3.63 x 10^3 km/s. So, because the electron was ejected at that speed, the energy required to eject it is added to its kinetic energy.

### Re: 1B. 15C

Posted: Sat Oct 19, 2019 9:32 pm
JamieVu_1D wrote:The velocity of an electron that is emitted from a metallic surface by a photon is 3.63 x 10^3 km/s.
(a) What is the wavelength of the ejected electron?
(b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 x 10^16 Hz. How much energy is
required to remove the electron from the metal surface?
(c) What is the wavelength of the radiation that caused photoejection of the electron?

For part a, I got 2.0 x 10^-10 m, and for part b I got 1.66 x 10^-17 J. In the solutions manual, the energy required to remove the electron is added to the electron's kinetic energy, and I don't understand why. Also, why can't I use the equation of lambda = h/mv?

Hi,

You cannot use the equation lambda=h/mv because that equation is for particles. When it asks about the wavelength of the radiation, it is asking about photons, in which you would use the equation E=hv. For part c, you are looking first for the energy of the photon required to remove the electron. For this, you would use the equation Ek=energy of photon- energy required to remove the electron. Rearranging the equation you get, energy of photon= kinetic energy (Ek) + energy required to remove the electron. Plug in your knowns and you get energy of photon= (1/2 x 9.109383 x 10^-31 x (3.63 x 10^3 km x 1000m/km) + 1.66 x 10^-17 = 2.25 x 10^-17 J. Since we know E=hv and c=lambda(v), we can now find wavelength by relating both equations with frequency. In c=lambda(v), when we rearrange we get v= c/lambda. Plugging that into E=hv, we get E=h(c/lambda). We can solve for lambda by rearranging to get lambda=(h)(c)/E = [(6.626 x 10^-34)(2.998 x 10^8)]/(2.25 x 10^-17) = 8.8 x 10^-9m = 8.8nm.

### Re: 1B. 15C

Posted: Sun Oct 20, 2019 5:55 pm
To respond to DHavo^^, I thought that you could use the two equations in similar ways because you can isolate and substitute v in both?

### Re: 1B. 15C

Posted: Wed Oct 30, 2019 8:10 pm
To find the wavelength you use DeBroglie's equation which is wavelength=planks constant/(massxvelocity) and when you plug in the given values the wavelength should come out to be 2.0x10-10m.