Chem Help!! It requires 183.7 kJ/mol to eject electrons from cesium metal.?
(a) What is the minimum frequency of light necessary to emit electrons from cesium via the photoelectric effect?_________s-1
(b)What is the wavelength of this light? ______nm
(c) If cesium is irradiated with light of 416 nm, what is the maximun possible kinetic energy of the emitted electrons? ______J
Please help..ive been working on this for a while, and cant seem to get the right answer!!
2006-10-26 15:28:43 · 3 answers · asked by lsubetty 2 in Education & Reference ➔ Homework Help
(a) E = hf
f = E / h
f = (183.7 * 10^3 J / 1 mol) / (6.6261 * 10^-34 J s) * (1 mol/ ...
6.02 * 10^23 atoms)
f = 4.605 x 10^14 s^-1
(b) lambda = c/f
lambda = (3 x 10^8 m/s) / (4.605 x 10^14 s^-1)
lambda = 651.42 nm
(c) E = hc/lambda
E = (6.6261 x 10^-34 J s) * (3 x 10^8 m/s) / (416 x 10^-9 m )
E = 4.78 x 10^-19 J
2006-10-26 15:39:33 · answer #1 · answered by Ryan 2 · 0⤊ 0⤋
Well I lost my chem textbook a couple of years ago, but I think there's a formula E=hv. You know E (183.7 kJ/mol) and h is a constant. Solve the equation to find the frequency, v.
To find wavelength, use the formula wavelength = 1/frequency
2006-10-26 22:31:25 · answer #2 · answered by Mary 3 · 0⤊ 0⤋
dang i know this stuff but too lazy to do...
first one, u divide 183. by avogadro's number, then plug into planck's equation, e=hv
solve for v to get answer for number one
number two, plug into c=v times sigma
number 3 you do it all backwards
sorry a bit lazy right now
2006-10-26 22:33:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋