Calculate the threshold energy for the nt Th232 → Th²³1 + 2n 231
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- Can you please help me with this question? Thank you so much!39.8 An atom of uranium-235 (atomic mass: m(23U) cays to thorium-231 (atomic mass: 231. 00 Th) = 231.03630 u) emission of an 235.04392 u) de- 92 via the a particle (nuclear 4.00151 u). m(ža) c = 2.998 x 108 m s-1. mass: Use %D (a) What is the binding energy of an 235U nucleus (in J)? 92 (b) What is the binding energy of a 231 Th nucleus (in J)? 90 (c) What is the binding energy of an a particle (in J)? (d) What is the maximum possible ki- netic energy (in J) of the a particle emitted during this decay? (Hint: what is the difference between the mass of 23°U and the total mass of 92 231 Th and an a particle?) 90 (e) Whatis the maximum velocity of the emitted a particle?Find the binding energy per nucleon in J and eV for the three following isotopes, given their atomic masses in u: 33 Si silicon 32.97800017 u ☐ × 10-12 J ☐ MeV 241 Es einsteinium 241.0685424 u 99 ×10−12 J☐MeV 170 oxygen 16.9991317012 u ×10-¹² J MeV mhydrogen = 1.007825 u mneutron 1.008664915 u -27 u = 1.6605 × 101 kg
- In class I derived the ordinary nuclear density to be about 0.138 u/fm3. A neutron star is a collapsed star that contains neutrons in a highly compactified state, so its average density is higher. Assume a neutron star is spherical and has an average density which is about twice the ordinary nuclear density. If it is 50% heavier than the Sun, what would be its radius? (Given: mass of sun = 2*1030 kg, 1 u = 1.66*10-27 kg.) A) 14.6 km B) 11.6 km C) 10.1 km D) none of these.1) For each of the following reactions work out the fastest interaction through which the conservation laws allow it to proceed. Explain your answers. If the reaction is forbidden by all interactions explain why: -+ a. pn++μ+ +μ¯ b. Aºp+e¯ c. μ΄ →e + V d. p+p→y+Y e. Kºn++ π + π° +7° f. π+pAº + Kº g. Aºn+p h. pnº+e+ + ve i. n→p+e+V₂Given: Nc = (2.51x1019)(mn/mo)3/2 (T/300)3/2 Nv = (2.51x1019)(mp/mo)3/2 (T/300)3/2 and ni = (NcNv)1/2 e(-Eg/2kT) show that: ni = (2.51x1019) ((mn/mo) * (mp/mo) )3/2 e(-Eg/2kT)