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A horizontal photoelectric plate whose work function is 1.9 eV is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity of ejected photoelectrons is non-positive? |
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Answer» Solution :MINIMUM energy of the photon falling on photoelectric plate `=hv=(hc)/(LAMDA)=((6.625xx10^(-34))(3xx10^(8)))/(250xx10^(-19))` ACCORDING to Einstein's photoelectric equation, `(1)/(2)mv^(2)=(hc)/(lamda)-W` `[v_(min)]=(2)/(m)[(hc)/(lamda)-W]` `=(2)/(9.1xx10^(-31))[((6.625xx10^(-34))(3xx10^(8)))/(250xx10^(-9))-1.9xx(1.6xx10^(19))]` `=1.08xx10^(12)` `v_(min)=1.04xx10^(6)ms^(-1)` |
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