A) \[\frac{hc}{2\lambda }\]
B) \[\frac{hc}{\lambda }\]
C) \[\frac{hc}{3\lambda }\]
D) \[\frac{3hc}{\lambda }\]
Correct Answer: A
Solution :
[a] From Einstein's photoelectric equation \[K.E{{.}_{\lambda }}=\frac{hc}{\lambda }-\phi \] ?.(i) (for monochromatic light of wavelength\[\lambda \]) where \[\phi \] is work function \[K.E{{.}_{\lambda /2}}=\frac{hc}{\lambda /2}-\phi \] ...(ii) (for monochromatic light of wavelength \[\lambda /2\]) From question, \[K.E{{.}_{\lambda /2}}=3\left( K.E{{.}_{\lambda }} \right)\Rightarrow \frac{hc}{\lambda /2}-\phi =3\left( \frac{hc}{\lambda }-\phi \right)\] \[\frac{2hc}{\lambda }-\phi =3\frac{hc}{\lambda }-3\phi \Rightarrow 2\phi =\frac{hc}{\lambda }\therefore \phi =\frac{hc}{2\lambda }\]
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