Question

The max velocity of the emitted electron is ________.,if the light of wave length 4300A is incident on the surface of potassium whoes work funtion is 2.3eV-> help please

Answer 1

The Einstein equation is

\[ K=h\nu - \Phi\]
where K is the kinetic energy of an emitted electron, h nu is the energy of a photon (Planck's constant times the frequency of a photon), and phi is the work function.

K can be rewritten as
\[K=\frac{1}{2}mv^2\]
so that you can solve directly for the velocity.

Does that help?

Answer 2

Remember that
\[1 A = 1 x 10^{-10}m\]
and that the frequency and wavelength of a photon are related by
\[c=\nu \lambda\]
\[ \nu = \text{ frequency}\]
\[ c = \text{ speed of light} = 3.0x10^8m/s\]
\[ \lambda= \text{ wavelength in m}\]
\[ 1eV = 1.602x10^{-19}J\]

Answer 3

so the velocity of the electron would be
\[v=\sqrt{2\frac{h \nu - \Phi}{m_e}}\]

\[h \propto \ m^2 \ kg/s \]
\[\Phi \propto J=kg \ m^2/s^2\]
\[m_e = \text{ mass of an electron}\]

Answer 4

I guess the first thing I should have asked is if you knew what a work function was :P Sorry :(

Answer 5

oooppss i guess i know ......any ways thanks for the help ... really apreciate it ....

Answer 6

Does that make sense?

Answer 7

@shkrina ?

Answer 8

Or did I just blow up a Greek alphabet and nonsense fell out?

Answer 9

sir .. thanks for ur help .. i did not get ur above .. question .. waht did not make sense
for u .. well the above statement says was i know .. that procedure which ugave me .. and in my text ist diffrent .. soo just wanted 2 clarify .. ..any more questions ..sir..^_^

Answer 10

@AllTehMaffs

Answer 11

I just couldn't tell what your ellipses meant is all ^_^ If you understand it then I don't have any questions; I just wanted to make sure!

Answer 12

(Curse you typos!) @shkrina He can't see your replies or tags 'cause right now the notifications system that tells you if you've been tagged is down ^^

Answer 13

thnku guys .. cyaaa ..

Answer 14

ciao ^_^

Answer 15

Oh, don't thank me @shkrina, I did nothing ^^
Just close the question and all is done (;
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Well, and give @AllTehMaffs a medal for his help.