If Molybdenum is irradiated with light of wavelength of 120nm,what is the maximu

Question

If Molybdenum is irradiated with light of wavelength of 120nm,what is the maximum possible kinetic energy of the emittedelectrons? so far I have E= h/ E=(6.626x10-34 )(3.00x108m/s)/1.20x10-7m E=1.66x10-18J The problem is that I don't know how to find theenergy/electron so that is where I am stuck. My book providesthe answer which is 9.3x10-19J/electron. Theproblem is I don't know how to arrive at that answer!!!! If Molybdenum is irradiated with light of wavelength of 120nm,what is the maximum possible kinetic energy of the emittedelectrons? so far I have E= h/ E=(6.626x10-34 )(3.00x108m/s)/1.20x10-7m E=1.66x10-18J The problem is that I don't know how to find theenergy/electron so that is where I am stuck. My book providesthe answer which is 9.3x10-19J/electron. Theproblem is I don't know how to arrive at that answer!!!!

Explanation / Answer

      Molybdenum metal must absorb radiation with aminimum frequency of 1.09 * 1015 s - 1before it can emit an electron from its surface via thephotoelectric effect.    Threshold frequency (v0)=1.09×1015 /s=1.09 ×1015 Hz    Wavelenth of light ()=120 nm=120×10-9 m=1.2 × 10-7
   Therefore,    Frequency (v)=c/= (3×108)/(1.2 × 10-7)=2.5 ×1015 Hz

According to Einstein's photoelectric equation,

Maxiumum kinetic energy of the emittedphotoelectron(Emax)=hv-hv0, where Emax is the maximumkinetic energy, v is The frequency of the incident radiation,v0 is the threshold frequency for the metal surface, andh is the Planck's constant.

Emax=hv-hv0         =h(v-v0)         =6.626068×10-34 (2.5 × 1015 -1.09×1015)          =9.34275588× 10-19 J         =h(v-v0)         =6.626068×10-34 (2.5 × 1015 -1.09×1015)          =9.34275588× 10-19 J

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