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Physics 19 Online
OpenStudy (anonymous):

The work function of cesium is 1.96 eV. If radiation of wavelength 4.00 × 10^2 nm is incident on the surface, find the kinetic energy of the ejected photoelectrons in eV and the speed of the ejected electrons.

OpenStudy (shamim):

hc/lamda=work funtion+0.5*mv^2

OpenStudy (shamim):

Here h=planks's constant Lamda=4*10^2nm=4*10^2*10^-6m Work function=1.96ev=1.96*1.6*10^-19J

OpenStudy (shamim):

c=light velocity m=mass of electron=9.1*10^-31kg v=?

OpenStudy (shamim):

Response plz!!!!

OpenStudy (anonymous):

I don't know the velocity

OpenStudy (shamim):

Frm my given equation can u try to find out v=?

OpenStudy (shamim):

Plz feel free to ask for clearing ur confusion!!!!

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