CHEM
posted by K .
I just have a couple of questions that I need some help with:
1) A 0.22caliber handgun fires a 27g bullet at a velocity of 765m/s.
Calculate the de Broglie wavelength of the bullet.
Express your answer using two significant figures in meters.
~and~
2) An Xray photon of wavelength 0.970nm strikes a surface. The emitted electron has a kinetic energy of 956eV.
What is the binding energy of the electron in kJ/mol?
KE=1/2mv^2; 1 electron volt (eV)=1.602*10^19J
Answer= BEmol = _______kJ/mol
please help if possible.

Didn't we work a De Broglie wavelength problem for you last night? What do you think you should do or do you have an answer you would like for us to check?

I'm not annonymous. I don't know how that name got there.

I'll get it right eventually.

the answer came to 3.2*10^35m. I kept typing the values into the calculator wrong. I multiplied the 765m/s, rather than dividing.
I still don't understand the other question though. Q#2. 
#1 is ok.
#2. I assume binding energy is the same as work function.
K. E. = (hc/lambda)  work function
Convert 956 ev to Joules and substitute for K. E.
h, c, and lambda (don't forget lambda is in meters). Solve for work function.
Check my thinking. Check for typos. 
Soooo???
KE = hc/lambda  1.602*10^19J*956eV??
6.6261*10^34*3.0*10^8 / 9.7*10^10m minus 1.53*10^16??
=5.19*10^17??? That can' be right?? 
I have 5.18 instead of 5.19.

apparently 5.18*10^17 kJ/mol isn't the correct answer either.

where am I going wrong?

The answer 5.18 x 10^17 is J/photon. You will need to change that to kJ/mol since the problem wants kJ/mol. Let me know if that isn't ok.

so just divide by 1000? I tried that when I had 5.19*10^20 & it told me I was incorrect?
or did you mean to divide somewhere else along the way? 
Divide by 1000 (1 kJ/1000 J) and multiply by 6.02 x 10^23 (6.02 x 10^23 photons/mol).

WOOO! 3.12*10^4 came to be correct indeed! Thank you ever so much for your time & patience.

So binding energy in your problem IS the same as work function which is the assumption I made in my first post. I've not heard it called that.

And thanks for letting me know.