The Cyclotron n order.to make a nuclear weapon out of uranium, you need a certai
ID: 2034463 • Letter: T
Question
The Cyclotron n order.to make a nuclear weapon out of uranium, you need a certain amount of "good" uranium. Uranium is more abundant as Uranium 235, but Uranium 238 is considered "good" for fission. To separate the "good" and "bad" Uranium it is ionized and fired into a chamber with a strong magnetic field. The different Uranium hits different places on a target plate and is then scraped off B-field (into the page) Target plate V-?MV1 X X X X X Uranium ion entering chamber The numbers 235 and 238 refers to how many neutrons and protons are present in each atom of Uranium 92 protons and either 143 neutrons or 146 neutrons respectively). The Uranium is ionized (a single electron is removed from the Uranium atoms). The mass of a Uranium 238 ion is 3.983x1025 kg. a) What is the mass of a Uranium 235 ion? These ionized atoms are accelerated with an electric potential drop of V 1 Megavolt. b) What is the charge of a Uranium ion (consider what it is missing)? c What is the speed at which the Uranium 235 exits the electric potential? d) What is the speed at which the Uranium 238 exits the electric potential? After being accelerated, they are fired into a chamber with a constant magnetic field of B 1.5 T (see above figue). There is a target plate positioned right next to where the ions enter the chamber. e) f) At what position do the U235 ions hit the plate? At what position do the U238 ions hit the plate?Explanation / Answer
a) Net force on charge 2, Fnet = k*q^2/d^2 + k*q^2/d^2
= 2*k*q^2/d^2
= 2*9*10^9*(9*10^-6)^2/(4*10^-3)^2
= 9.11*10^4 N
b) Fnet = 0
c) Fnet = sqrt(2)*k*q^2/d^2
= sqrt(2)9*10^9*(9*10^-6)^2/(4*10^-3)^2
= 6.44*10^4 N
a) mass of Uranium-235, m = (235/238)*3.983*10^-25
= 3.933*10^-25 kg
b) q = 1.602*10^-19 C
c) we know, KE = (1/2)*m*v^2
==> v = sqrt(2*KE/m)
= sqrt(2*1*10^6*1.6*10^-19/(3.933*10^-25))
= 9.02*10^5 m/s
d) v = sqrt(2*KE/m)
= sqrt(2*1*10^6*1.6*10^-19/(3.983*10^-25))
= 8.96*10^5 m/s
e) d_235 = 2*r_235
= 2*m_235*v/(B*q)
= 2*3.933*10^-25*9.02*10^5/(1.5*1.6*10^-19)
= 2.956 m
f) d_238 = 2*r_238
= 2*m_238*v/(B*q)
= 2*3.983*10^-25*8.96*10^5/(1.5*1.6*10^-19)
= 2.974 m
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