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A 3310 Å photon liberates an electron from a material with energy 3 xx 10^(-19) J while another 5000 Å photon ejects an electron with energy 0.972 xx 10^(-19) J from the same material. Determine the value of Planck's constant and the threshold wavelength of the material. |
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Answer» SOLUTION :To find `:` Planck's constant `H = upsilon_(0)` `h upsilon_(1) = phi + K.E_(1) , h upsilon_(1) rarr` Energy of photon 1 `h rarr` Plank's constant `phi rarr `Work funcltion `K.E_(1) rarr `Kinetic energy of an electron 1 `upsilon_(1) rarr` Frequency of photon 1 For photon 2, `h upsilon_(2) = phi + K.E_(2)` Given `:` `lambda_(1) = 3310Å = 3310 xx 10^(-10)m ` and `lambda_(2) = 5000 xx 10^(-10) m ` `h upsilon_(1) = ( h^(@) C )/( lambda_(1)) , K.E_(1) = 3xx 10^(-19) J, K.E_(2) = 0.972 xx 10^(-19) J ` `( hc)/( 3310 xx 10^(-10)) = phi + 3 xx 10^(-10)`...(1) `( hC )/( 5000 xx 10^(-10)) = phi + 0.972 xx 10^(-19) `...(2) `hc = 3310 xx 10^(-10) phi + 9930 xx 10^(-29) `.....(3) (2) can be rewritten as `hc = 5000 xx 10^(-10) phi+ 0972 xx 5000 xx 10^(-29)`....(4) From ( 3) `hc = 3310 xx 10^(-10) phi + 9930 xx 10^(-29)` `- 3310 xx 10^(10) [hc] = +- 16,086,600 xx 10^(-39)` `5000 xx 10^(-10) [hc = +49,650,000xx10^(-39)` `1690 xx 10^(10) hc = 33,563,4000 xx 10^(-39)` `rArr h=(33,563,400 xx 10^(-39))/(1690 xx 3 xx 10^(8)) = 6,620 xx 10^(-31)` `h = 6.62 xx 10^(-34)` Solving equation (3) and ( 4) `( 5000 - 3310 ) xx 10^(-10) phi = 5070 xx 10^(-29) = 0` `phi = ( 5070 xx 10^(-29))/( 1690 xx 10^(-10)) , phi = 3 xx 10^(-19)` `( hc)/(lambda_(0)) = 3 xx 10^(-19)` `lambda_(0) = ( 6.62 xx 10^(-34) xx 3 xx 10^(8))/( 3 xx 10^(-19)) = 6620 xx 10^(-10)` `= 6620 =6.62 xx 10^(-7) Js= 6620 xx 10^(-10) m ` |
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