(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]
Verified Correct Answer: .
Physical Principles & Numerical Solution
(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
Source: Myschool Past Questions Archive (Record #54129)