Light of wavelength 5000 A falls o n a metal surface of work function 1.9 eV . Find (a ) the energy of photons (b) kinetic energy of photoelectrons . [1eV = 1.6 * 10 (-19) J ] please answer fast

Here we will use the relation

E = KE + hν_{0} (1)

where E is the energy of the incident photon, KE is the energy of the emitted photon and hν_{0} is the work function or the minimum energy required an electron from the surface of the metal. Now as

E = hν = (hc) / λ

where h is the planck's constant, c is the speed of light, λ is its wavelength of radiation and ν is its frequency. Thus the energy of the incident photon is

E = (6.63 X 10^{-34} X 3 X 10^{-8}) / (5000 X 10^{-10})

= 3.978 X 10^{-19} J

Now 1 eV = 1.6 X 10^{-19} J

So 1 J = 1 / (1.6 X 10^{-19}) eV

Therefore 3.978 X 10^{-19} J = (1 X 3.978 X 10^{-19} ) / (1.6 X 10^{-19})

= 2.486 eV

Hence the energy of the photon is 2.486 eV.

Now we are given that the work function is 1.9 eV. Substituting the above value and the value of work function in equation (1), we get

KE = E - hν_{0}

= 2.486 - 1.9

= 0.586 eV

So the kinetic energy of the emitted photon is 0.586 eV.

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