Consider a heavy nucleus initially at rest in the laboratory frame. Suppose it u

Question

Consider a heavy nucleus initially at rest in the laboratory frame. Suppose it undergoes spontaneous fission into two identical nuclei. The rest mass of the heavy nucleus is 2.0000 times 10^-25 kg and the energy liberated in the process (the total kinetic energy of the fission products) is 5.4000 times 10^-10 J. What is the rest mass of each of the resulting nuclei? What is the speed of each of the resulting nuclei? consider 10 times ionized 23 11 Na. Use the Bohr theory to Calculate the ground state energy of the remaining electron. Calculate the speed on the ground state electro Calculate the orbital radius of this ion. Consider the spring-mass system depicted below. The spring constant is 100 N/m. The mass is 0.250 kg. The friction is such as to cause the amplitude to decay exponentially with the number of oscillations according to: The position of the mass ranges from-A to +A (delta x = 2A). The momentum of the mass ranges from -p_max to +p_max (Delat px = 2p_maa) where p_max can be determined by conservation of energy If A_0 = 0.100 m, how many osci ations will the system execute (before no further oscillations would be detectable)? Ka x-ray radiation is emitted when an electron that has been removed from an atom's K-shell (n = 1) is replaced by an electron from the L-shell (n = 2). An approximate value for the Ka wavelength may be calculated using the Bohr theory for hydrogenic atoms by neglecting all electron-electron interactions except for the "screening" effect of the K-shell. After one of the K-shell electrons has been removed, there is only one electron left in the K-shell. The "screening" effect of this electron effectively reduces the atomic number of the atom by one (Z_eff = Z - 1). Apply the above considerations, and use R_infinity in the equations to determine the approximate wavelength of the K alpha radiation for 64/29 Cu, R_infinity 1.097 times 10^7 m^-l

Explanation / Answer

2] Let the mass of heavy nuclei be M and daughter nuclei be m, Balancing mass

Mc^2 = 2 mc^2 + Energy

2e-25*3e8^2 = 2*m*3e8^2 + 5.4e-10

1.8e-8-0.054e-8 = 2*m*3e8^2

m = [1.8e-8-0.054e-8]/18e16

= 9.7*10^-26 kg answer

b] let speed be v,

2*0.5mv^2 = Energy

v = sqrt( 5.4e-10/9.7e-26 )

= 74612400 m/s

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