use N(/theta)d/theta = 2I(/theta)e((/theta)*(/theta))/((/theta)*(/theta)))d((/theta)))/((/theta)*(/theta))
45) An alpha particle (charge 2e) is aimed directly at a gold nucleus (charge 79e). What minimum initial kinetic energy must the alpha particle have to approach within 5.0 * 10-14 m of the center of the gold nucleus before reversing direction? Assume that the gold nucleus, which has about 50 times the mass of an alpha particle, remains at rest.
Answer is:
4.6 MeV
Explanation:
Related Atoms and Nuclei MCQ with Answers
Answer is:
3.63 MeV
Explanation:
kinetic energy is the negative of total energy. K.E. = -E also P.E. is equal to negative of two time of total energy. P.E.=-2E
Answer is:
3.63 MeV
Explanation:
kinetic energy is the negative of total energy. K.E. = -E also P.E. is equal to negative of two time of total energy. P.E.=-2E
Answer is:
-218 eV, 16 times
Explanation:
kinetic energy is the negative of total energy. K.E. = -E also P.E. is equal to negative of two time of total energy. P.E.=-2E
Answer is:
218 eV, 16 times
Explanation:
kinetic energy is the negative of total energy. K.E. = -E also P.E. is equal to negative of two time of total energy. P.E.=-2E