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A photoelectric surface is illuminated successively by monochromatic light of wavelength lambda and (lambda)/(2).If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times in the first case,the work function of the surface of the material is: (h=Plank's constant ,c=speed of light) |
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Answer» `(HC)/(3lambda)` In first case ,`K_(1)=(hc)/(lambda)-phi` In second case ,`K_(2)=(hc)/((lambda)/(2))-phi=(2hc)/(lambda)-phi` But `K_(2)=3K_(1)` is given . `THEREFORE (2hc)/(lambda)-phi=(3hc)/(lambda)-3phi` `3phi-phi=(3hc)/(lambda)-(2hc)/(lambda)` `2phi=(hc)/(lambda)` `therefore phi=(hc)/(2lambda)` |
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