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Photonof wavelength4xx 10^(7)strikes onmetalsurfacethe workfunctionof the metal being2.13 eVCalculate(i) Theenergyof the photonev(ii)The kineticenergyof theemissionand(iii) thevelocityof thephotoelectron . (1 eV = 1.6020 xx 10^(19) J ) |
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Answer» Solution :`1 EV= 1.6020 XX 10 ^(19)|` where `lambda = 4 xx 10^(7) m` `=(2.13 eVxx 1.6020 xx 10^(19)J)/( 1eV ) ` (i) Calculateof energyof photon `E=- (hc)/( lambda ) = (6.626 xx 10^(34)J s) (3.0 xx 10^(8)ms^(-1))/(4XX 10^(7) m)` (ii) Calculationof kineticenergyof emission Kineticenergyof emission= (photonenergyworkfunctionW ) `=(4.9695xx 10^(19) - 3.4123 xx 10^(19)J` (iii) Calculation ofvelocityof thephotoelectronkineticenergy`1.56 xx 10^(19) J` Kineticenergy`=(1)/(2) MV^(2)` `v^(2)= (2xx "kineticenergy")/( m )` `=sqrt((2xx1.56 xx 10^(19))/(9.1 xx 10^(31) kg ))` `= sqrt (0.3429 xx 10^(12) kg^(-1))` `=0.5855 xx 10^(6) ms^(-1) ` `5.855 xx 10^(5) ms^(-1)` |
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