Electrons are emitted from the surface of a metal when it\'s exposed to light. T

Question

Electrons are emitted from the surface of a metal when it's exposed to light. This is called the Photoelectric effect. Each metal has a certain threshold frequency of light below which nothing happens Right at this threshold frequency, an electron is emitted Above this frequency, the electron is emitted and the extra energy is transferred to the electron. The equation for this phenomenon is KE = hv - hV_0 where KE is the kinetic energy of the emitted electron, h = 6.63 Times 10^-34 J.B is planck's constant, v is the frequency of the light and v_0 is the threshold frequency of the metal. Also, since E = hv, the equation can also be written as KE = E - E_0 where K is the energy of the light and E_0 is the threshold energy of the metal. Here are some data collected on a sample of cesium exposed to various energies of light What is the threshold frequency V_0 of cesium? Note that 1ev (electron volt) + 1.60 Times 10^-19 J. Express your answer numerically in hertz. V_0 = Hz

Explanation / Answer

From the given formula:

Light energy (h*v) = Threshold energy (h*vo) +K.E

=> h*vo = 3.91-0.02 =3.89 eV

=> vo = 3.89 eV/(4.135667662(25)×1015 eV.s) =9.4059797e+14 Hz

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