(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are pr...
Question 1 Report
(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]
Answer Details
(a) Transmutation equation
Nitrogen-14 bombarded by a neutron emits an alpha particle:
\[ {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n} \;\rightarrow\; {}^{4}_{2}\text{He} + {}^{11}_{5}\text{B} \](Mass numbers: \(14+1 = 4+11\); atomic numbers: \(7+0 = 2+5\). The residual nucleus is boron-11.)
(b)(i) A radioactive atom has an unstable nucleus that spontaneously disintegrates, emitting radiation (alpha, beta or gamma), whereas a stable atom has a nucleus that does not decay.
(b)(ii) Half life is the time taken for half the atoms (or nuclei) in a given sample of a radioactive material to decay.
(b)(iii) \(105\,\text{days} = 3\times 35\,\text{days} = 3\) half lives. Fraction remaining:
\[ \left(\tfrac{1}{2}\right)^3 = \frac{1}{8} \](c) Photoelectric calculations (\(\lambda = 5.00\times10^{-7}\,\text{m}\), \(W_0 = 1.90\,\text{eV}\)):
(i) Photon energy:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{5.0\times10^{-7}} = 3.96\times10^{-19}\,\text{J} \ (\approx 2.48\,\text{eV}) \](ii) Maximum kinetic energy: work function \(W_0 = 1.90\times1.6\times10^{-19} = 3.04\times10^{-19}\,\text{J}\).
\[ E_k = E - W_0 = 3.96\times10^{-19} - 3.04\times10^{-19} = 9.2\times10^{-20}\,\text{J} \ (0.575\,\text{eV}) \](iii) Stopping potential:
\[ V_s = \frac{E_k}{e} = \frac{9.2\times10^{-20}}{1.6\times10^{-19}} = 0.575\,\text{V} \]Download The App On Google Playstore
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