The photoelectric effect occurs when the frequency of incident radiation is above a threshold frequency. If we double this incident frequency, what effect will it have on the experiment? Also what will happen if we double the wavelength?

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    $\begingroup$ Please take the time to use proper grammar and capitalization in your questions. I fixed this one for you, but in the future please do it yourself. It is a courtesy to the people take time to answer your questions $\endgroup$ – DanielSank Jul 8 '15 at 21:53

The energy of the photon used to reap the electron from the atom and give it some kinetic energy, thus -
$h\nu =\frac{hc}{\lambda}=\frac{1}{2}mv^2+W_0$, where $W_0$, the energy needed to reap the electron from the atom, is constant per type of atom. Thus $ v=\sqrt{\frac{2(h\nu-W_0)}{m}}=\sqrt{\frac{2(\frac{hc}{\lambda}-W_0)}{m}}$

As long as $h\nu=\frac{hc}{\lambda}>W_0$ - the electrons will be reaped from the metal. Changing frequency/wavelength of the photons won't change the number of reaped electrons, but will change their speed/kinetic energy.

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    $\begingroup$ For completeness (last part of the question), if you double the wavelength you halve the energy. If you were just above the cutoff, you will now be below it and there will be no electrons emitted. You might want to edit that into your answer. $\endgroup$ – Floris Jul 8 '15 at 22:24
  • $\begingroup$ It's embedded in the $h\nu=\frac{hc}{\lambda}>W_0$ condition $\endgroup$ – Alexander Jul 8 '15 at 23:37
  • $\begingroup$ I realize that, but I stand by my recommendation to be explicit. "Abundantly clear" is a good strategy. $\endgroup$ – Floris Jul 9 '15 at 0:25

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