PHYSICS FOR SCIEN & ENGNR W/MOD MAST
4th Edition
ISBN: 9780134112039
Author: GIANCOLI
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 37, Problem 81GP
To determine
The closeness of
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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.
85
83
85
83
Determine the disintegration energy (Q-value) in MeV.
Determine the binding energy (in MeV) for 211 At.
85
EB =
From 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 remaining
Chapter 37 Solutions
PHYSICS FOR SCIEN & ENGNR W/MOD MAST
Ch. 37.2 - Prob. 1AECh. 37.2 - Prob. 1BECh. 37.4 - Prob. 1CECh. 37.7 - Prob. 1DECh. 37.7 - Prob. 1EECh. 37.11 - Prob. 1FECh. 37 - Prob. 1QCh. 37 - Prob. 2QCh. 37 - Prob. 3QCh. 37 - Prob. 4Q
Ch. 37 - Prob. 5QCh. 37 - Prob. 6QCh. 37 - Prob. 7QCh. 37 - Prob. 8QCh. 37 - Prob. 9QCh. 37 - Prob. 10QCh. 37 - Prob. 11QCh. 37 - Prob. 12QCh. 37 - Prob. 13QCh. 37 - Prob. 14QCh. 37 - Prob. 15QCh. 37 - Prob. 16QCh. 37 - Prob. 17QCh. 37 - Prob. 18QCh. 37 - Prob. 19QCh. 37 - Prob. 20QCh. 37 - Prob. 21QCh. 37 - Prob. 22QCh. 37 - Prob. 23QCh. 37 - Prob. 24QCh. 37 - Prob. 25QCh. 37 - Prob. 26QCh. 37 - Prob. 27QCh. 37 - Prob. 28QCh. 37 - Prob. 1PCh. 37 - Prob. 2PCh. 37 - Prob. 3PCh. 37 - Prob. 4PCh. 37 - Prob. 5PCh. 37 - Prob. 6PCh. 37 - Prob. 7PCh. 37 - Prob. 8PCh. 37 - Prob. 9PCh. 37 - Prob. 10PCh. 37 - Prob. 11PCh. 37 - Prob. 12PCh. 37 - Prob. 13PCh. 37 - Prob. 14PCh. 37 - Prob. 15PCh. 37 - Prob. 16PCh. 37 - Prob. 17PCh. 37 - Prob. 18PCh. 37 - Prob. 19PCh. 37 - Prob. 20PCh. 37 - Prob. 21PCh. 37 - Prob. 22PCh. 37 - Prob. 23PCh. 37 - Prob. 24PCh. 37 - Prob. 25PCh. 37 - Prob. 26PCh. 37 - Prob. 27PCh. 37 - Prob. 28PCh. 37 - Prob. 29PCh. 37 - Prob. 30PCh. 37 - Prob. 31PCh. 37 - Prob. 32PCh. 37 - Prob. 33PCh. 37 - Prob. 34PCh. 37 - Prob. 35PCh. 37 - Prob. 36PCh. 37 - Prob. 37PCh. 37 - Prob. 38PCh. 37 - Prob. 39PCh. 37 - Prob. 40PCh. 37 - Prob. 41PCh. 37 - Prob. 42PCh. 37 - Prob. 43PCh. 37 - Prob. 44PCh. 37 - Prob. 45PCh. 37 - Prob. 46PCh. 37 - Prob. 47PCh. 37 - Prob. 48PCh. 37 - Prob. 49PCh. 37 - Prob. 50PCh. 37 - Prob. 51PCh. 37 - Prob. 52PCh. 37 - Prob. 53PCh. 37 - Prob. 54PCh. 37 - Prob. 55PCh. 37 - Prob. 56PCh. 37 - Prob. 57PCh. 37 - Prob. 58PCh. 37 - Prob. 59PCh. 37 - Prob. 60PCh. 37 - Prob. 61PCh. 37 - Prob. 62PCh. 37 - Prob. 63PCh. 37 - Prob. 64PCh. 37 - Prob. 65PCh. 37 - Prob. 66PCh. 37 - Prob. 67PCh. 37 - Prob. 68PCh. 37 - Prob. 69PCh. 37 - Prob. 70PCh. 37 - Prob. 71PCh. 37 - Prob. 72GPCh. 37 - Prob. 73GPCh. 37 - Prob. 74GPCh. 37 - Prob. 75GPCh. 37 - Prob. 76GPCh. 37 - Prob. 77GPCh. 37 - Prob. 78GPCh. 37 - Prob. 79GPCh. 37 - Prob. 80GPCh. 37 - Prob. 81GPCh. 37 - Prob. 82GPCh. 37 - Prob. 83GPCh. 37 - Prob. 84GPCh. 37 - Prob. 85GPCh. 37 - Prob. 86GPCh. 37 - Prob. 87GPCh. 37 - Prob. 88GPCh. 37 - Prob. 89GPCh. 37 - Prob. 90GPCh. 37 - Prob. 91GPCh. 37 - Prob. 92GPCh. 37 - Prob. 93GPCh. 37 - Show that the wavelength of a particle of mass m...Ch. 37 - Prob. 95GPCh. 37 - Prob. 96GPCh. 37 - Prob. 97GPCh. 37 - Prob. 98GPCh. 37 - Prob. 99GPCh. 37 - Prob. 100GP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- Explain how a hydrogen atom in the ground state (l = 0) can interact magnetically with an external magnetic field.arrow_forwardAlpha 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.arrow_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
- Someone 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_forwardIn 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_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
- Which hydrogen like atom has the wavelength difference between the first line of Lyman and Balmer is equal to 59.3nm?arrow_forwardThe atomic mass of hypothetical element (voll aic) (₁4A) is 4.02 amu. The energy required to dissociate this element into its component parts is: (m₁ =1.009 amu, mp = 1.007amu, neglect the mass of electron)arrow_forwardEach a particle in a beam of a particles has a kinetic energy of 4.5 MeV. Through what potential difference would you have to accelerate these a particles in order that they would have enough energy so that if one is fired head-on at a gold nucleus it could reach a point 1.9 10^-14 m from the center of the nucleus?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning