I just have a couple of questions that I need some help with:
1) A 0.22-caliber 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 X-ray 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??