Electrons are ejeched from a certain metallic surface with a

Electrons are ejeched from a certain metallic surface (with a work function of 2.46 ev) with a speed of 1 times 10^6 m/s. Find the wavelength of the lenght that is skriking the surface. The half life of an isotope of phosphrus is 14 days. If a souiple contains 3.0 times 10^16 nuclei, determine its activity. Express your answer in curies After 2 days, the activity of a rodisachive sauiple drops to 84.2% of its original activity. Calculate its half - life. After a plant or an animal dios, its 14_C contents decrease with a half life of $730 years. if an archeologiot finds an aucient firepit confoiningo partially consumed firwood and the 14_C is only 12.5% that of an equal

Solution

KE of ejected electron = m v^2 / 2

     = (9.109 x 10^-31 ) (1 x 10^6)^2 / 2

   = 4.5545 x 10^-19

KE ( in eV) = 2.85 eV

energy of photon = work function + KE of electron

      = 2.46 + 2.85 = 5.31 eV

and energy = h c / lambda

lambda = 1240 eV-nm / 5.31 = 233.67 nm

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activity = lambda * N

where lambda = ln(2) / T = ln(2) / (14 x 24 x 3600 sec)

            = 5.73 x 10^-7 per sec

activity = 3 x 10^16 x 5.72 x 10^-7 = 1.719 x 10^10 nuclei per sec

in Curie, = 1.719 x 10^10 / (3.7 x 10^10) = 0.465 Ci

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N = No 2^(-t/T)

0.842N0 = N0 2^(-t/T)

-2/T = ln(0.842) / ln(2)

T = 2/0.248= 8.06 days

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0.125 No = No 2(-t/5730)

-t/5730 = ln(0.125)/ln(2)

t = 17190 years