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In a photoelectric effect experimetent, a metallic surface of work function 2.2 eV is illuminated with a light of wavelenght 400 nm. Assume that an electron makes two collisions before being emitted and in each collision 10% additional energy is lost. Q. After how many collisions will the electron be unable to come out of the metal?A. 2B. 6C. 4D. 8 |
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Answer» Correct Answer - C Energy of photon `E=(hc)/(lamda)=(1.24xx10^(3))/(400)=3.1 eV` Remaining energy `=3.1-0.31=2.79eV` Energy lost is first collision is `(3.1)xx((10)/(100))=0.31eV` Remaining energy is `3.1-0.31=2.79 eV` Energy lost in second collision is `(2.79)xx((10)/(100))=0.279 eV` Total energy lost two collisions is `(0.31)+(0.279)eV=0.589 eV` So, from conservation of energy, we have `(hc)/(lamda)=phi+KE_(max)+` energy lost in two collision `3.1=2.2+KE_(max)+0.589` `KE_(max)=0.31eV` Total energy after second collision is `(2.79-0.279)=2.511 eV` Energy lost in third collision is `2.511xx(10)/(100)=0.2511 eV` Remaining energy `=(2.511-0.2155)=2.2500 eV` Energy lost in fourth collision `=(2.2599xx(10)/(100))=0.2259 eV` Remaining energy `=(2.2599-0.2250)=2.034 eV` After the fourth collision, the electron doen not have enough energy to overcome the work function, so it cannot come out. |
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