# Maximum velocity of an electron emitted from lanthanum

• lpettigrew
In summary, the conversation discusses finding the maximum kinetic energy of an electron and using the photoelectric equation. The formula used is 1/2mv^2 max = E - Φ, where Φ is the work function of lanthanum. However, the units were not consistent, leading to incorrect calculations. After correcting the units and converting eV to J, the final speed of the electron is calculated to be 1.3 x 10^6 ms^-1.
lpettigrew
Homework Statement
I have found the multiple choice question below but cannot deliberate upon a solution, and am sure there is a problem with my workings. Could anyone help me here?

What is the maximum velocity of an electron emitted from lanthanum under a light of wavelength 150 nm?
A. 1.3 x 10^-3 ms-1
B. 1.3 x 10^3 ms-1
C. 1.3 x 10^6 ms-1
D. There is not sufficient energy to emit an electron.
Relevant Equations
1/2 mv^2 max = hf - Φ
E=hf
Mass of an electron = 9.1*10^-31
Well, to find the maximum kinetic energy of the electron use E=hf, E=hc/λ=6.63*10^-34*3*10^8/1.5 *10^-7
E=1.326 *10^-18 J
1/2m v^2 max=E

Rearrange in terms of v:
v=√2E/m
v=√2*1.326 *10^-18/ *9.1*10^-31
v=1707127... ~ 171000 ms^-1

Where have I gone wrong here?

How difficult is it to bump an electron out of Lanthanum ?

lpettigrew
What is the work function of lanthanum?
You have written:
1/2 mv^2 max = hf - Φ and
1/2m v^2 max = E.
So which one is it?

lpettigrew
Steve4Physics said:
What is the work function of lanthanum?
You have written:
1/2 mv^2 max = hf - Φ and
1/2m v^2 max = E.
So which one is it?
I think it would be the first formula, which is the photoelectric equation. Sorry but I am still struggling here.

lpettigrew said:
I think it would be the first formula, which is the photoelectric equation. Sorry but I am still struggling here.

I asked " What is the work function of lanthanum? " but you haven't answered this.

Yes, it is the first formula. But you have used the second formula!

You used
1/2m v^2 max=E
but (to spell it out when I shouldn't) you should have used
1/2m v^2 max=E - Φ

Look up Φ for lanthanum if not already given the value, convert to joules if needed and use it!

By the way writing " Mass of an electron = 9.1*10^-31" is a bad mistake - units missing!

Last edited:
Steve4Physics said:
I asked " What is the work function of lanthanum? " but you haven't answered this.

Yes, it is the first formula. But you have used the second formula!

You used
1/2m v^2 max=E
but (to spell it out when I shouldn't) you should have used
1/2m v^2 max=E - Φ

Look up Φ for lanthanum if not already given the value, convert to joules if needed and use it!

By the way writing " Mass of an electron = 9.1*10^-31" is a bad mistake - units missing!
Thank you for your reply. I thought you meant find the work function of lanthanum through calculation. This is 3.5 eV. Should I also convert the energy to eV: 1.326 *10^-18 = 8.2762414 ~ 8.3 eV
Sorry, Mass of an electron = 9.1*10^-31 kg
Using, 1/2m v^2 max=E - Φ then;
Rearrange in terms of v;
v=√2(E - Φ)/m
v=√2*(8.3-3.5)/9.1*10^-31 kg
v=3.24799 * 10^15 ~ 3.25 *10^15 ms^1

I think I am wrong again though!

Check the units of each of your factors !

I used eV for E and Φ and kg for mass. Would this be wrong?

Yes -- if you want to get m/s you have to convert the eV to kg m2/s2, a.k.a. Joules

lpettigrew
BvU said:
Yes -- if you want to get m/s you have to convert the eV to kg m2/s2, a.k.a. Joules
Oh, ok.
E=1.326 *10^-18
Φ=3.5 eV~5.6*10^-19 J

v=√2(E - Φ)/m
v=√2*(1.326 *10^-1.326 *10^-18)/9.1*10^-31 kg
v=1297503.944 ~ 1.3 x 10^6 ms^-1 (option c)
Right, thank you for your help

BvU
lpettigrew said:
I used eV for E and Φ and kg for mass. Would this be wrong?

BvU beat me to it, but I'll add:

Using ½mv² for energy with m in kg and v in m/s means your energy unit is the joule (J) (standard SI units)

Work functions are often given in energy units of eV. So you must convert eV to J (1eV = 1.6x10⁻¹⁹J).

Well done for sticking with it and sorting it out! We learn best that way.

lpettigrew
Steve4Physics said:
BvU beat me to it, but I'll add:

Using ½mv² for energy with m in kg and v in m/s means your energy unit is the joule (J) (standard SI units)

Work functions are often given in energy units of eV. So you must convert eV to J (1eV = 1.6x10⁻¹⁹J).

Well done for sticking with it and sorting it out! We learn best that way.
Thank you very much for your help and guidance

## 1. What is the maximum velocity of an electron emitted from lanthanum?

The maximum velocity of an electron emitted from lanthanum is dependent on the energy of the incident photon, the work function of lanthanum, and the properties of the material's surface.

## 2. How is the maximum velocity of an electron emitted from lanthanum measured?

The maximum velocity of an electron emitted from lanthanum can be measured using various techniques such as photoelectron spectroscopy, Auger electron spectroscopy, and scanning electron microscopy.

## 3. What factors affect the maximum velocity of an electron emitted from lanthanum?

The maximum velocity of an electron emitted from lanthanum is affected by the energy of the incident photon, the work function of lanthanum, and the properties of the material's surface such as crystal structure and defects.

## 4. How does the maximum velocity of an electron emitted from lanthanum differ from other elements?

The maximum velocity of an electron emitted from lanthanum may differ from other elements due to differences in their work function, surface properties, and electronic structure.

## 5. What applications does the knowledge of the maximum velocity of an electron emitted from lanthanum have?

The knowledge of the maximum velocity of an electron emitted from lanthanum is important in various fields such as materials science, surface chemistry, and semiconductor technology. It can also aid in the understanding of the electronic properties of lanthanum and its compounds.

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