what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays...

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency of 1.2 x 10^15 hz? Express your answer in joules to three significant figures.

Solutions

Expert Solution

The kinetic energy of the emitted electrons can be calculated using the formula:

K = E - where is the work function of the concerned metal

For Cesium, = 2.1 eV = 2.1 * 1.6e-19 J = 3.36e-19 J

Now, energy of incident light, E = h where h is the planck constant and is the frequency of incident light.

Hence,

K = E -

= 6.626e-34 * 1.2e15 - 3.36e-19 = 7.95e-19 - 3.36e-19 = 4.59e-19 J

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queen_honey_blossom answered 1 year ago

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