Answer to Question #112533 in Electricity and Magnetism for zuwi

Question #112533
the work function of silver is 4.73eV EM radiation with a frequency of 1.2x10^15 Hz strikes a piece of pure silver,what is the speed of the electrons that are emitted
1
Expert's answer
2020-04-28T09:34:59-0400

The photoelectric effect equation says

hf=ϕ+mv22hf=\phi+\frac{mv^2}{2}

Hence, the speed of the electrons that are emitted

v=2(hfϕ)mv=\sqrt{\frac{2(hf-\phi)}{m}}

=2(6.67×1034×1.2×10154.73×1.6×1019)9.1×1031=\sqrt{\frac{2(6.67\times 10^{-34}\times 1.2\times 10^{15}-4.73\times 1.6\times 10^{-19})}{9.1\times 10^{-31}}}

=3.1×105m/s=3.1\times 10^{5}\:\rm m/s

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment