Wavelength of radiation

[Chem_Mod]

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6x10^3 km/s. What is the wavelength of the radiation that caused photoejection of the electron?" From part a and b, we have found that the wavelength of the ejected electron is 2.02x10^-10 m and that the energy required to remove the electron from the metal surface is 1.66x 10^-17 J. The problem also states that the frequency of the radiation is 2.50x 10^16 Hz. I don't understand why I can not use the equation c= wavelength x frequency to solve for the wavelength, using the frequency 2.50 x 1-^16.

[Chem_Mod]

In this case, a photon hits a metal surface and not only ejects an electron but also makes it "fly" off with a certain velocity. This means that the energy of the incoming photon exceeded the energy required to eject the electron. The problem states that no electrons can be ejected until the photon has a frequency of AT LEAST 2.50 x10^16 Hz (this can be converted to the work function of the metal by using the equation E=hv). But we already know the photon had more energy than that because the electron was not only ejected but went off with some velocity. In this case, we need to use the conservation of energy equation, E(incoming photon)= (phi, work function of the metal)+(kinetic energy of the outgoing electron, 1/2mv^2, where m is the mass of the electron and v is the velocity in m/sec) in order to find the total energy of the photon, and then we can determine its wavelength. By using the work function energy you determined in part b and the velocity of the electron stated in the problem, you can plug in E(incoming photon)= (1.66 x10^-17J) + (1/2)*(9.10939 x10^-31 kg)*(3.6 x10^6 m/sec)^2 = 2.25 x10^-17 J. That is the energy of the incoming photon. Now we can go back to our energy of light equation to solve for frequency, E=hv, or v= (2.25 x10-17 J)/(6.626 x10^-34 J*sec)= 3.3957 x10^16 Hz. And lastly solve for wavelength using the speed of light equation, (lambda)=(3.0 x10^8 m/sec)/(3.3957 x10^16 Hz)= 8.8 x10^-9 m.

[804584179]

I still don't understand part c. So after you find the energy required to remove the electron from the metal surface, how would you find the wavelength? I'm assuming you'd use k=1/2mv^2 and then after that which equation?

[KNguyen_1I]

In part C, you use that 1/2mv^2 to find the kinetic energy of the electron ejected. But, to find the energy of the incoming photon, you must add that energy to the energy threshold. Why? Conceptually this implies that the energy of the incoming photon was enough to overcome the threshold energy and therefore must have an energy equal or greater to the threshold energy, and as the electron emitted has a velocity we know it has kinetic energy.

In physics, the sum of kinetic and work energy basically gives you the total input energy (eg the energy of the photon). Once you have that, then you can fanangle with the numbers to get the wavelength of the incoming photon.

Hope this helps!

[Camille 4I]

What equation would we use to relate velocity with wavelength?

[brennayoung]

In the initial question, wouldn't you need to convert the velocity of the electron 3.6x10^3km.s^-1 into m.s^-1 so it would be 3.6m.s^-1 to be in SI units or am I mistaken?

[Nan_Guan_1L]

In the initial question, wouldn't you need to convert the velocity of the electron 3.6x10^3km.s^-1 into m.s^-1 so it would be 3.6m.s^-1 to be in SI units or am I mistaken?

for part a of this question, yes you would need to convert that to SI units, but it's 3.6x10^6 m/s. the value should be getting bigger because 1 m/s is smaller unit than 1 km/s. hope that helps!

[Nan_Guan_1L]

Just want to share:
for part c, I was really confused whether you could use the 2.50 x 10^6 hz from part b because that was not given in the main question. but it turned out that the problem can't be solved just using what's given in the main question. Also, when I plugged in 9.11 x 10^-31 for mass of electron, and got 2.84 x 10^-17 for total energy. and finally I got 7.00 x 10^9 as the answer. My point is even though I kept three significant figures, my answer was still pretty off, which made me confused for a very long time..：）

[arielle_cunanan3K]

How did you do part a? Relating the wavelength to the velocity

[sophiavmr]

For part A you would use the De Broglie equation for wavelength, so wavelength= h/p or h/mv.

[Sally_Luo_3F]

In the initial question, wouldn't you need to convert the velocity of the electron 3.6x10^3km.s^-1 into m.s^-1 so it would be 3.6m.s^-1 to be in SI units or am I mistaken?

for this part it would be 1km=1000m so the conversion would be 3.6*10^6 m*s^-1

[Anna Turk 1D]

For part A you would use the De Broglie equation for wavelength, so wavelength= h/p or h/mv.

Why would we need to use the de Broglie equation and can't use the lambda=c/v equation instead?

[605537776]

I have the same question as Anna. Why are we using that equation over c = λf.