Maximum velocity of an electron emitted from lanthanum

[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?

 

[BvU]

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

 

[Steve4Physics]

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]

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.

 

[Steve4Physics]

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!

 

[lpettigrew]

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!

 

[BvU]

Check the units of each of your factors !

 

[lpettigrew]

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

 

[BvU]

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

 

[lpettigrew]

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

 

[Steve4Physics]

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]

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