It takes 4.6 eV to remove one of the least tightly bound electrons from a metal surface. When monochromatic photons strike the metal surface, electrons having kinetic energy from zero to 2.2 eV are ej

Question

It takes 4.6 eV to remove one of the least tightly bound electrons from a metal surface. When monochromatic photons strike the metal surface, electrons having kinetic energy from zero to 2.2 eV are ejected. What is the energy of the incident photons?

TestHub · Accepted Answer

According to question, the work function of the metal surface is $W = 4.6 eV$ and $K . E$ of ejected electron = $2.2 eV$

Let the energy of the incident photon is $E$ .

i.e. $E = W + K_{e}$

$K_{e} =$ the kinetic energy of ejected electrons

$E = 4. 6 eV + 2 .2 eV$

$= 6. 8 eV$

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