Can someone explain in detail the steps for thisproblem? The binding energy for

Question

Can someone explain in detail the steps for thisproblem?   
The binding energy for an electron on a surface of metallicsodium is 3.9 x 10-19J. Find the kinetic energy of thephotoelectrons when sodium is illuminated with UV light of 305nm. Answer: 2.62 x 10-19 J Can someone explain in detail the steps for thisproblem?   
The binding energy for an electron on a surface of metallicsodium is 3.9 x 10-19J. Find the kinetic energy of thephotoelectrons when sodium is illuminated with UV light of 305nm. Answer: 2.62 x 10-19 J

Explanation / Answer

so B.E. = 3.9*10^(-19)J(given) Now metallic sodium illuminated with light of wavelength =305nm So energy associated with that light can be found as      E =hc/ where h=planck'sconstant=6.63*10^(-34)    c=velocity of light       =wavelength of light putting the values we get E    E=6.516964*10^(-19)J therefore the kinetic energy of the ejected electron can begiven as       K.E. =energy of incident photon- B.E.       so the answer comes out to be=2.616964*10^(-19)J

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