## HW 1.33

ErinKim1I
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am

### HW 1.33

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^3 km/s. Part B says: 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?

I'm thinking I probably have to use the work function somewhere, but how do I go about this problem?

Thanks

Tarek Abushamma
Posts: 30
Joined: Fri Apr 06, 2018 11:01 am

### Re: HW 1.33

You can use this given frequency to calculate the threshold energy using the equation, E = (h)(v). From there you should be able to use the work function to calculate the energy of the photon needed to displace an electron the given 3.6x10^3 km/s.

Anna De Schutter - 1A
Posts: 66
Joined: Wed Feb 21, 2018 3:01 am

### Re: HW 1.33

Hi!

The information given to us in this question is that no electron is emitted until the frequency of radiation reaches 2.50*10^16 Hz. We are trying to find the amount of energy required to remove the electron from the surface, thus the threshold energy (also known as work function) required to remove an electron from the surface. When given this kind of question where you need to find the work function you set Ek (kinetic energy of the photon) equal to 0 m/s because you are looking for the threshold energy needed to remove the electron.

We know:
Ep (energy of photon) = work function + Ek (kinetic energy of electron)

So:
work function = Ep - Ek
Because Ek=0m/s:
work function = Ep = h*frequency = (6.626*10^-34 Js)*(2.50*10^16 s-1)=1.66*10^-17 J

So the final answer is: the energy required to remove an electron from the surface equals 1.7*10^-17 J.

I hope this helps! :)
Anna De Schutter - section 1A

Joanna Pham - 2D
Posts: 113
Joined: Fri Apr 06, 2018 11:04 am

### Re: HW 1.33

Could someone explain how to do part A please? Would I just use the equation lambda = c/(frequency), with the frequency being E/h? If so, would E in this case be the kinetic energy of the electron?

Posts: 59
Joined: Fri Apr 06, 2018 11:04 am

### Re: HW 1.33

For part A, you can use the equation lambda=hc/Energy of the electron or (1/2mv^2. Then, you plug in Planck's constant and the speed of light on the top and the mass of an electron and the given velocity(3.6x10^6 when converted from km/s to m/s) on the bottom. The answer would be 33.7 nm, so the wavelength of the ejected electron is 33.7 nm.

Tiffany Chen 1A
Posts: 36
Joined: Fri Apr 06, 2018 11:02 am
Been upvoted: 1 time

### Re: HW 1.33

Since you're given the velocity of an electron, to do part A, you can use the De Broglie equation lambda = h/mV where m is the mass of an election (you can just look this up, it's 9.11x10^-31) and V is the velocity and then solve for the wavelength.

EllenRenskoff-1C
Posts: 32
Joined: Fri Apr 06, 2018 11:04 am

### Re: HW 1.33

Regarding part D, I got that the wavelength was 8.8 nm from part C. Would this be considered part of the x ray and gamma ray radiation type even though the book says that's for wavelengths less than 3 nm? Is it because it its closer to that value than to the next type of radiation?

Tarek Abushamma
Posts: 30
Joined: Fri Apr 06, 2018 11:01 am

### Re: HW 1.33

8.8nm would still be a part of the X-ray spectrum, I believe.

Maya Khoury
Posts: 29
Joined: Fri Apr 06, 2018 11:03 am

### Re: HW 1.33

Where does the value for the frequency come from? I tried finding the frequency using the c = Λv but I don't seem to get the value 2.5 x 10¹⁶s⁻¹

Bianca Nguyen 1B
Posts: 36
Joined: Fri Apr 06, 2018 11:04 am

### Re: HW 1.33

The frequency 2.5 x 10^16 Hz was given to you in part b of the question

Chem_Mod
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### Re: HW 1.33

The first statement in this problem goes with part c). The frequency given in b) is only used to get the work function.

Bree Perkins 1E
Posts: 34
Joined: Fri Apr 06, 2018 11:04 am

### Re: HW 1.33

Anna De Schutter - 1A wrote:Hi!

The information given to us in this question is that no electron is emitted until the frequency of radiation reaches 2.50*10^16 Hz. We are trying to find the amount of energy required to remove the electron from the surface, thus the threshold energy (also known as work function) required to remove an electron from the surface. When given this kind of question where you need to find the work function you set Ek (kinetic energy of the photon) equal to 0 m/s because you are looking for the threshold energy needed to remove the electron.

We know:
Ep (energy of photon) = work function + Ek (kinetic energy of electron)

So:
work function = Ep - Ek
Because Ek=0m/s:
work function = Ep = h*frequency = (6.626*10^-34 Js)*(2.50*10^16 s-1)=1.66*10^-17 J

So the final answer is: the energy required to remove an electron from the surface equals 1.7*10^-17 J.

I hope this helps! :)
Anna De Schutter - section 1A

My question is why is Ek=0?

Anna De Schutter - 1A
Posts: 66
Joined: Wed Feb 21, 2018 3:01 am

### Re: HW 1.33

Ek=0m/s in this question (part b) because the question is asking us to find "how much energy is required to remove the electron from the metal surface". This implies that we are looking for the threshold energy (also called the work function) in this question and the threshold energy means the amount of energy needed to remove an electron from a metal surface without any kinetic energy. Afterwards, if the energy of the photon exceeds this threshold energy, the electron will have kinetic energy but when we are looking for the threshold energy, we know Ek=0m/s.

I hope this helps! :)
Anna De Schutter - section 1A