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### 1B 15 7th edition

Posted: Sat Oct 20, 2018 8:51 pm
15. Velocity of electron emitted from a metallic surface by a photon is 3.6 km/s. What is the wavelength of the ejected electron?

For this question, I tried to use the kinetic energy formula 1/2mv^2 but it doesn't work out. Why can't I use the KE formula and how do you know it's supposed to be De Broglie's equation that's used instead?

### Re: 1B 15 7th edition

Posted: Sat Oct 20, 2018 9:15 pm
You can't use KE for wavelength because you don't know the threshold energy to find the total energy.

### Re: 1B 15 7th edition

Posted: Sun Oct 21, 2018 11:45 am
Since we are unable to use the Ek formula, which formula must we use in order to solve the problem?

### Re: 1B 15 7th edition

Posted: Sun Oct 21, 2018 6:02 pm
You can use De Broglie's equation (wavelength=h/(m)(v)) since it relates wavelength to the momentum of the electron.

### Re: 1B 15 7th edition

Posted: Mon Oct 22, 2018 6:20 pm
For part c I keep getting 12nm and not 8.8nm (the answer). What am I doing wrong as I am plugging in the exact formula for speed of light?

### Re: 1B 15 7th edition

Posted: Mon Oct 22, 2018 8:53 pm
For part c I also keep getting 12nm, it might be a typo in the book.

### Re: 1B 15 7th edition part C

Posted: Tue Oct 23, 2018 5:14 pm
For 1B 15 c :
The question is: What is the wavelength of the radiation that caused photoejection of the electron? So first we have to find the Energy of this wavelength. We know the threshold energy from part 1B 15 b=1.66x10^-17.

We also know that:

Energy (per photon) - threshold energy =Ek

We can rearrange this to be: Energy = threshold energy + Ek. (Additionally Ek=1/2mv^2)

So we now have:

E=threshold energy + 1/2mv^2

E=1.66x10^-17 + 1/2(9.1095x10^-31)(3.6x10^6)^2

E= 2.25x10^-17

So now we know the energy per photon. Now we have to find the wavelength that corresponds to this energy.

E=ch/wavelength

We can rearrange this to be: wavelength= ch/E.

wavelength = ch/2.25x10^-17

wavelength= 8.82x10^-9 m

:)