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Ultraviolet light of wavelength 2271 Å f...

Ultraviolet light of wavelength 2271 Å fro a 100 W mercury source irradiates a photocell made of molybdenum metal. If the stopping potential is -1.3V, estimate the work function of the metal. How would the photocell respond to a high intensity `(~10^(5)Wm^(-2))` red light of wavelength 6328 Å produced bya He-Ne laser ?

Text Solution

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Here `lamda=2271Å=2271xx10^(-10)m=2.271xx10^(-7)m`, and stopping potential `V_(0)=-1.3V`
Using Einstein.s photoelectronic equation `hv-phi_(0)=eV_(0)`, we have
`phi_(0)=hv-eV_(0)=(hc)/(lamda)-eV_(0)=(6.63xx10^(-34)xx3xx10^(8))/(2.271xx10^(-7))-1.3xx1.6xx10^(-19)J`
`=8.758xx10^(-19)-2.08xx10^(-19)J=6.678xx10^(-19)J=(6.678xx10^(-19))/(1.6xx10^(-19))eV=4.17eV`
For He-Ne laser light of wavelength `lamda.=6328Å=6.328xx10^(-7)m,` the enrgy of a photon will be
`E.=(hc)/(lamda.)=(6.63xx10^(-34)xx3xx10^(8))/(6.328xx10^(-7))J=(6.63xx10^(-34)xx3xx10^(8))/(6.328xx10^(-7)xx1.6xx10^(-19))eV=1.964eV`
As `E.ltphi_(0)`, hence no photoelectric emission will take place irrespective of intenstiy of laser beam.
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