A metal ejects electrons with a velocity of 8.4 X 105 m/s when the energy of the incident photons is twice
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A metal ejects electrons with a velocity of 8.4 X 105 m/s when the energy of the incident photons is twice

[From: ] [author: ] [Date: 12-05-18] [Hit: ]
so you know the final KE of the electrons, (1/2)mv^2....OK,......
the energy of the work function. What’s the frequency of the incident photons?

Answer f = 9.7 X 1014 Hz

Could you please show me the steps to how to get the answer? Thanks!

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"A metal ejects electrons with a velocity of 8.4 X 10^5 m/s..."

OK, so you know the final KE of the electrons, (1/2)mv^2.

"... when the energy of the incident photons is twice the energy of the work function"

OK, so you know that half the energy of the photons went into overcoming the work function, and the other half is the KE.

So the photon energy hf is equal to 2 * the KE above. Put in the numbers and solve for f.
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