(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are produced. Represent this transmutation in a symbolic equation.
(b)(i) How does a radioactive atom differ from a stable one?
(ii) Explain ‘half life’.
(iii) A sample of radioactive material has a haft life of 35 days. Calculate the fraction of the original quantity that will remain after 105 days.
(c) Light of wavelength 5.00 x 10(^{-7})m is incident on a material of work function 1.90 eV. Calculate
(i) photon energy.
(ii) kinetic energy of the most energetic photo electron.
(iii) stopping potential [Plancks constant h =6.6 x 10(^{-34})Js] [c= 3.0 x10(^8)ms(^{-2}), leV= 1.6 x 10(^{19})J]
Explanation
(a) (^{14}_7 + ^1_0n) -> (^4_2He + ^{11}_5X)
(b)(i) The radioactive atom disintegrates spontaneously while a stable atom does not disintegrate on its own.
(ii) The time at which the number of radioactive atoms remaining is just one-half of the original number is called half-life of the element. The half-life T(_{frac{1}{2}}) is related to the decay constant ((lambda)) of the isotope by the equation.
(T frac{1}{2} = frac{0.693}{ lambda})
(iii) After 35days, 1/2 of it remains.
After 70 days, 1/2 of 1/2 i.e. 1/4 remains. After 105 days, % of 1/4 i.e. 1/8 remains.
(c)(i) The photon energy Ep = hf = (frac{hc}{f})
= (frac{6.6 times 10^{-34} x 3.0 times 10^8}{ 5.0 x 10^{-7}})
= 3.96x 10-19J
(ii) The kinetic energy E(_k) = (Ep – W)
E(_k) = hf – W = (frac{hc – W}{lambda})
W = 1.9eV = 1.6 x 10(^{19}) x 1.9
= 3.044 x 10(^{-19})J.
E(_k) = 3.96 x 10(^{-19}) – 3.044 x 10(^{19}) = 9.2 x 10(^{-20})J.
(iii) Stopping potential Vs = energy needed to stop most energetic electron.
eVs =E(_{kmax})
Vs = (frac{E_{max}}{e})
= (frac{9.20 times 10^{-20}}{1.6 times 10^{-19}})
= 0.575 V