Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780136139225
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 37, Problem 81GP
To determine
The closeness of
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In a Rutherford scattering experiment, an a-particle (charge = +2e) heads directly toward a gold nucleus (charge = +79e). The α-particle had a kinetic energy of 5.0 MeV when very far (r→ ∞) from the nucleus. Assuming the gold nucleus to be fixed in space, determine the distance of closest approach. Hint: Use conservation of energy with PE =kq1q2/r.
Use the below values for this problem. Please note that the mass for H is for the entire atom (proton & electron).
Neutron: m,= 1.67493x1027 kg= 1.008665 u = 939.57 MeVIC
H: my = 1.67353x10 27 kg = 1.007825 u = 938.78 MeVic
1u= 1.6605x10-27 kg = 931.5 MeVic?
Consider the following decay: 211 At 207 Bi + a. 211 At has a mass of 210.9874963 u, 207 Bi has a mass of 206.981593 u, and a has a mass of 4.002603 u.
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Determine the disintegration energy (Q-value) in MeV.
Determine the binding energy (in MeV) for 211 At.
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EB =
Alpha particles are projected toward a gold foil from a distance that is sufficiently large to consider the Coulomb force negligible. The gold nuclei have 118 neutrons and 79 protons. If a 3.45 MeV alpha particle has a scattering angle of 180° and the gold nucleus does not recoil, determine the distance of closest approach of the alpha particle.
Chapter 37 Solutions
Physics for Scientists and Engineers with Modern Physics
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- Explain how a hydrogen atom in the ground state (l = 0) can interact magnetically with an external magnetic field.arrow_forwardFrom far away a proton is fired directly toward the center of the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm. Assume the mercury nucleus is fully-ionized, with no electrons bound to it. If the proton is fired at a speed of 3.8×107 m/s, what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest. Express your answer with the appropriate units. ► View Available Hint(s) 0 d = 2.7 Submit A fm Previous Answers ? X Incorrect; Try Again; 13 attempts remainingarrow_forwardUse the below values for this problem. Please note that the mass for H is for the entire atom (proton & electron). Neutron: m = 1.67493x10-27 kg = 1.008665 u = 939.57 MeV/c² . ¹H: mH = 1.67353x10-27 kg = 1.007825 u = 938.78 MeV/c² 1 1 u = 1.6605x10-27 kg = 931.5 MeV/c² . Consider the following decay: 239 Pu 235 U+ a. 239 Pu has a mass of 239.0521634 u, 235 U has a mass of 235.0439299 u, and a has a mass of 4.002603 u. 94 92 94 92 Determine the disintegration energy (Q-value) in MeV. Q = Determine the binding energy (in MeV) for 239 Pu. 94 EB =arrow_forward
- The proton (mass M) and electron (mass m) in a hydrogen atom actually rotate about their common center of mass as shown in Figure . The distance r = re + rM is still defi ned to be the electron-nucleus distance. Show that Equation is only modifi ed by substituting for m byarrow_forwardThe first line in the Lyman series of H is at 82257.098 cm-1, while that line in the spectrum of deuterium (D) is at 82281.476 cm-1.Check the option that contains the mass of the D nucleus, calculated from this spectral information. Assume that h = 6,63 x 10-34 J.s, c = 2,99 x 108 m/s, m(elétron) = 9,11 x 10-31 kg e m(próton) = 1,67 x 10-27 kgarrow_forwardAn alpha particle with kinetic energy 11.0 Me V 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%=pob, where po is the magnitude of the initial momentum of the alpha particle and b=1.50x10-12m (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.) Repeat for b=1. 10×10-13 m. Express your answer in meters. ΑΣφ Submit Request Answer Part C Repeat for b=1.50×10-14 m. Express your answer in meters.arrow_forward
- In the experiment, alpha particles follow a hyperbolic trajectory. The figure shows an alpha particle A initially approaching a gold nucleus N and being deflected at an angle 0 = 90°. N is located at a focus of the hyperbola, and the trajectory of A passes through a vertex of the hyperbola. ++► x y = x y =-x a. Determine the equation of the trajectory of the alpha particle if d = 4 x 10 14 m. Choose the correct equation below. O A. X= -. + 16 x 10 - 28 В. X= Vy + 16 ×10-28 OC. D. X= + 4x 10 - 28 X = - 4x 10-28 b. What was the minimum distance between the centers of the alpha particle and the gold nucleus? (Use scientific notation. Round to the nearest tenth as needed.)arrow_forwardSomeone wanted to build a scale model of the atom with a nucleus 1.4 m in diameter. How far away would the nearest electron need to be in meters? Assume the orbital radius of an electron is 10^(-10)m, while the radius of the nucleus is 10^(-15) m. ra=?arrow_forwardUsing the Thomson model, estimate the average scatter- ing angle when alpha particles (z = 2) with kinetic energy 3 MeV are scattered from gold (Z = 79). The atomic radius of gold is 0.179 nm.arrow_forward
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