Question

If the alpha particles have an initial kinetic energy of 7.7 MeV, then assuming a head-collision...

If the alpha particles have an initial kinetic energy of 7.7 MeV, then assuming a head-collision between an alpha particle (helium nucleus with +2e charge) and a gold nucleus (79 protons, so +79e charge), and using conservation of energy at the point of closest approach when all of the alpha particle's kinetic energy is converted to electric potential energy, calculate the approximate distance of closest approach (and thus coarsely estimate the size of the nucleus)

0 0
Add a comment Improve this question Transcribed image text
Answer #1

ge + (+79 el let closest approach is r According to law of So conservation of energy 7.7 MeV = I (20) (798) UTE ४ 158e2 J UTE

Thanks for asking.

Add a comment
Know the answer?
Add Answer to:
If the alpha particles have an initial kinetic energy of 7.7 MeV, then assuming a head-collision...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • An alpha particle with a kinetic energy of 12.0 MeV makes a head-on collision with a...

    An alpha particle with a kinetic energy of 12.0 MeV makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. A gold nucleus has 79 protons, and an alpha particle is a helium nucleus consisting of two protons and two neutrons. The mass of an alpha particle is 6.64424 x 10-27 kg....

  • An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a...

    An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point change. The atomic number of gold is 79, and the alpha particle is a helium nucleus consisting of 2 protons and 2 neutrons).

  • In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider...

    In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy K heading directly for the nucleus of a gold atom (charge =+79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus for the case K = 3.5 MeV

  • An alpha particle with kinetic energy 13.0 MeV makes a collision with lead nucleus, but it...

    An alpha particle with kinetic energy 13.0 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.30×10−12 m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number...

  • An alpha particle with kinetic energy 11.0 MeV makes a collision with lead nucleus, but it...

    An alpha particle with kinetic energy 11.0 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.50×10−12 m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number...

  • An alpha particle with kinetic energy 12.5 MeV makes a collision with lead nucleus, but it...

    An alpha particle with kinetic energy 12.5 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.50×10−12 m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number...

  • 4th time asking this. Very pissed at the "experts" here. Please be correct An alpha particle...

    4th time asking this. Very pissed at the "experts" here. Please be correct An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10.3 MeV (10.3 times 10^6 EV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Answer the following questions about the collision. What is the initial momentum...

  • In 1910 Rutherford performed a classic experiment in which he directed a beam of alpha particles...

    In 1910 Rutherford performed a classic experiment in which he directed a beam of alpha particles at a thin gold foil. He unexpectedly observed a few of the particles scattered almost directly backward. This result was not consistent with then current models of atomic structure and led Rutherford to propose the existence of a very dense concentration of positive charge at the center of an atom—the atomic nucleus. The alpha particle has a charge of +2e and the gold nucleus...

  • Problem 10.23 An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out...

    Problem 10.23 An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 9.9 MeV (9.9 x 100 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Answer the following questions about the collision. What is the initial momentum of the alpha particle? (You may assume its speed is small...

  • An alpha particle with kinetic energy 12.0 MeVmakes a collision with lead nucleus, but it is...

    An alpha particle with kinetic energy 12.0 MeVmakes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.20×10−13 m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT