What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.4×1015Hz?

Question

anonymous · Answer

This is my work so far.

aaronq · Answer

What's the ionization energy for the electrons emitted?

aaronq · Answer

$\sf KE=E-\phi =h\nu -I.E.$

anonymous · Answer

Light energy 
(eV)	Electron emitted?	Electron KE 
(eV)
3.87	no	—
3.88	no	—
3.89	yes	0
3.90	yes	0.01
3.91	yes	0.02

Here's the data table I was given for Cesium. I found the threshold frequency for Cesium to be 9.39*10^14 Hz. I'm just not sure how to solve this problem and where I'm going wrong, @aaronq

aaronq · Answer

Okay, the threshold frequency is also called the work function, $\phi$.

The energy of a photon is used to remove the electron, the leftover is the kinetic energy.

$\sf E_{photon }=KE+\phi\rightarrow    KE=E_{photon}-\phi$

$ \sf KE=E_{photon}-\phi=h\nu -h\nu_o=h(\nu-\nu_o)$

$\sf KE=(6.626*10^{-34}J*s)(1.4*10^{15}Hz-9.39*10^{14} Hz)$

anonymous · Answer

Okay, that makes much more sense. Thank you for explaining it to me, I should be able to work it out from here!

aaronq · Answer

okay cool! no problem