How much kinetic energy must an alpha particle have before its distance of closest approach to a gold nucleus is equal to the nuclear radius (7.0 x 10-15 m)?
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How much kinetic energy must an alpha particle have before its distance of closest approach to a gold nucleus is equal to the nuclear radius (7.0 x 10-15 m)?
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- Iodine-131 is a beta-emitter, which decay by the first-order rate law with a half-life of 8.0 days. What is the activity in cps (counts per second) of 25.0 ng (2.5 x 10–8 g) sample of 131I (atomic mass = 131 g/mol)? (Avogadro’s #, NA = 6.02 x 1023/mol) (A) 1.2 x 108 cps (B) 1.5 x 1010 cps (C) 1.0 x 1011 cps (D) 4.1 x 1011 cpsIs the following statement true? "An electron travels along multiple paths from its emission by a beta-decaying nucleus to a Geiger counter where it is detected"Use the uncertainty principle to estimate the minimum speed and kinetic energy of an alpha particle confined to the interior of a heavy nucleus.
- -14 (a) Find the speed an alpha particle requires to come within 3.4 x 10 m of an iron nucleus. 19.35*10**6 X Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the energy of the alpha particle in MeV. 7.79 Your response differs from the correct answer by more than 100%. MeVIn beta decay, the energy of the electron or positron emitted from the nucleus lies somewhere in a relatively large range of possibilities. In alpha decay, however, the alpha particle energy can only have discrete values. Why is there this difference?A living specimen in equilibrium with the atmosphere contains one atom of 14C (half- life = 5 730 yr) for every 7.7 x 1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22011) contains 27 mg of carbon. When the sample is placed inside a shielded beta counter with 89% counting efficiency, 837 counts are accumulated in one week. Assuming that the cosmic-ray flux and the Earth's atmosphere have not changed appreciably since the sample was formed, find the age of the sample in the nearest year (1 yr ~ 3.1536 × 10' s). Take Avogadro Number to be DONT ROUND IN THE STEPS BECAUSE IT WILL NavG =6.026x1023 AFFECT YOUR ANSWER AND MAKE IT WRONG
- Taking into account the recoil (kinetic energy) of the daughter nucleus, calculate the kinetic energy K(alpha) of the alpha particle in the following decay of a 238U nucleus at rest. 238U⟶234Th+alpha K(alpha) = ? MeVHow much kinetic energy must an alpha particle have before its distance of closest approach to a gold nucleus is equal to the nuclear radius (7.0 fm)An α-particle moving with initial kinetic energy K towards a nucleus of atomic number z approaches a distance ‘d’ at which it reverses its direction. Obtain the expression for the distance of closest approach ‘d’ in terms of the kinetic energy of α-particle K.
- A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life = 5 730 yr) for every 7.7 × 1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22O11) contains 21.0 mg of carbon. When the sample is placed inside a shielded beta counter, 837 counts are accumulated in one week. Assuming that the cosmic-ray flux and the Earth’s atmosphere have not changed appreciably since the sample was formed, find the age of the sample. ( Atomic mass of C is 12u)In a real or imaginary nucleus of 45X⁹7, (a) how many protons are in the nucleus, (b) how many neutrons are in the nucleus, and (c) how many electrons are in orbit about the nucleus, assuming the atom is electrically neutral? (a) Number (b) Number i (c) Number Units Units UnitsA wooden artifact is found in an ancient tomb. Its carbon-14 (14/6C) activity is measured to be 60.0% of that in a fresh sample of wood from the same region. Assuming the same amount of 14C was initially present in the artifact as is now contained in the fresh sample, determine the age of the artifact.