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2001 WAEC Physics Theory (a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons,...

Physics
WAEC 2001

(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]

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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


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