Explanation Given Info : Copper atoms have 29 protons in their nuclei and the copper nucleus is taken as a sphere with a diameter of 4.8 × 10 − 15 m . The charge on the alpha particle is equal to + 2 e . The nucleus of copper is a sphere. Alpha particle is present at rest at infinity. To bring this alpha particle from rest at infinity to the surface of the nucleus, certain amount of work is required. Therefore, the amount of work required to bring an alpha particle from rest at infinity to the surface of the nucleus is equal to the potential energy possessed by the alpha particle. The formula to calculate the net amount of work is, W = V × q (1) Here, W is the amount of work. V is the electrical potential. q is the charge present on alpha particle. The electrical potential used in above formula is to be calculated. Copper atoms comprises of 29 protons. Therefore, the charge present on copper nucleus is equal to 29 e . The diameter of copper nucleus is 4.8 × 10 − 15 m . The formula to calculate the radius of copper nucleus is, Radius = Diameter 2 Substitute, 4.8 × 10 − 15 m for diameter to find radius of the copper nucleus. Radius = Diameter 2 = 4.8 × 10 − 15 m 2 = 2.4 × 10 − 15 m Thus, the calculated value of radius of copper nucleus is 2.4 × 10 − 15 m . The formula to calculate the electrical potential is, V = k × Q r Here, k is the proportionality constant and is equal to 9 × 10 9 Nm 2 /C 2 . Q is the charge present on copper nucleus and is equal to 29 e . r is the radius of copper nucleus and is equal to 2.4 × 10 − 15 m . Substitute, 9 × 10 9 Nm 2 /C 2 for k , 29 e for Q , 2.4 × 10 − 15 m for r to find electrical potential of the copper nucleus. V = k × Q r = 9 × 10 9 Nm 2 /C 2 × 29 e 2

5th Edition

ISBN: 8220103026918

Author: Walker

Publisher: PEARSON

See similar textbooks

Concept explainers

Rutherford's scattering experiment

Rutherford’s scattering experiment is also known as the gold foil experiment. This experiment disproved the plum pudding model of J.J. Thomson. In a gold foil experiment, high-energy alpha particles are made to strike a thin gold foil. The scattered alph…

Question

Chapter 31, Problem 4PCE

To determine

The amount of work required to bring an alpha particle (charge =+2 e ) from rest at infinity to the “surface” of the nucleus.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 31, Problem 3PCE

Chapter 31, Problem 5PCE

Students have asked these similar questions

Starting from the Clasius-Clapeyron equation, Lo + constant kBT log p and assuming the latent heat of evaporation per gram of water to be constant, and equal to L= 2260 J/g , find the boiling temperature of water in degrees Celcius at a pressure of 50.0 atm. Each oxygen atom has 8 protons and 8 neutrons in its nucleus. The boiling temperature of water at a pressure of 50.0 atm is around 250 °C. How do you account for the difference? R = 8.31 //(mol · K) , m,-1.67x1027 kg , NA=6.02x1023 mol-1

An alpha particle (Z = 2, mass = 6.64 x 10-27 kg) approaches to within 1.00 x 10-14 m of a carbon nucleus (Z = 6). What are (a) the maximum Coulomb force on the alpha particle, (b) the acceleration of the alpha particle at this time, and (c) the potential energy of the alpha particle at the same time?

Use 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 =

Chapter 31 Solutions

EBK PHYSICS

Ch. 31.1 - Prob. 1EYU Ch. 31.2 - Prob. 2EYU Ch. 31.3 - Prob. 3EYU Ch. 31.4 - Prob. 4EYU Ch. 31.5 - Prob. 5EYU Ch. 31.6 - Prob. 6EYU Ch. 31.7 - Prob. 7EYU Ch. 31 - Prob. 1CQ Ch. 31 - Prob. 2CQ Ch. 31 - Prob. 3CQ

Ch. 31 - Prob. 4CQ Ch. 31 - Prob. 5CQ Ch. 31 - Prob. 6CQ Ch. 31 - Prob. 7CQ Ch. 31 - Prob. 8CQ Ch. 31 - Prob. 9CQ Ch. 31 - Prob. 1PCE Ch. 31 - Prob. 2PCE Ch. 31 - Prob. 3PCE Ch. 31 - Prob. 4PCE Ch. 31 - Prob. 5PCE Ch. 31 - Prob. 6PCE Ch. 31 - Prob. 7PCE Ch. 31 - Prob. 8PCE Ch. 31 - Prob. 9PCE Ch. 31 - Prob. 10PCE Ch. 31 - Prob. 11PCE Ch. 31 - Prob. 12PCE Ch. 31 - Prob. 13PCE Ch. 31 - Prob. 14PCE Ch. 31 - Prob. 15PCE Ch. 31 - Prob. 16PCE Ch. 31 - Prob. 17PCE Ch. 31 - Prob. 18PCE Ch. 31 - Prob. 19PCE Ch. 31 - Prob. 20PCE Ch. 31 - Prob. 21PCE Ch. 31 - Prob. 22PCE Ch. 31 - Prob. 23PCE Ch. 31 - Prob. 24PCE Ch. 31 - Prob. 25PCE Ch. 31 - Prob. 26PCE Ch. 31 - Prob. 27PCE Ch. 31 - Prob. 28PCE Ch. 31 - Prob. 29PCE Ch. 31 - Prob. 30PCE Ch. 31 - Prob. 31PCE Ch. 31 - Prob. 32PCE Ch. 31 - Prob. 33PCE Ch. 31 - Prob. 34PCE Ch. 31 - Prob. 35PCE Ch. 31 - Prob. 36PCE Ch. 31 - Prob. 37PCE Ch. 31 - Prob. 38PCE Ch. 31 - Prob. 39PCE Ch. 31 - Prob. 40PCE Ch. 31 - Prob. 41PCE Ch. 31 - Prob. 42PCE Ch. 31 - Prob. 43PCE Ch. 31 - Prob. 44PCE Ch. 31 - Prob. 45PCE Ch. 31 - Prob. 46PCE Ch. 31 - Prob. 47PCE Ch. 31 - Prob. 48PCE Ch. 31 - Prob. 49PCE Ch. 31 - Prob. 50PCE Ch. 31 - Prob. 51PCE Ch. 31 - Prob. 52PCE Ch. 31 - Give the electronic configuration for the ground...Ch. 31 - Prob. 54PCE Ch. 31 - Prob. 55PCE Ch. 31 - Prob. 56PCE Ch. 31 - The configuration of the outer electrons in Ni is...Ch. 31 - Prob. 58PCE Ch. 31 - Prob. 59PCE Ch. 31 - Prob. 60PCE Ch. 31 - Prob. 61PCE Ch. 31 - Prob. 62PCE Ch. 31 - Prob. 63PCE Ch. 31 - Prob. 64PCE Ch. 31 - Prob. 65PCE Ch. 31 - Prob. 66PCE Ch. 31 - Prob. 67PCE Ch. 31 - Prob. 68GP Ch. 31 - Prob. 69GP Ch. 31 - Prob. 70GP Ch. 31 - Prob. 71GP Ch. 31 - Prob. 72GP Ch. 31 - Prob. 73GP Ch. 31 - Prob. 74GP Ch. 31 - Prob. 75GP Ch. 31 - Prob. 76GP Ch. 31 - Prob. 77GP Ch. 31 - Prob. 78GP Ch. 31 - Prob. 79GP Ch. 31 - Prob. 80GP Ch. 31 - Prob. 81GP Ch. 31 - Prob. 82GP Ch. 31 - Prob. 83GP Ch. 31 - Prob. 84PP Ch. 31 - Prob. 85PP Ch. 31 - Prob. 86PP Ch. 31 - Prob. 87PP Ch. 31 - Prob. 88PP Ch. 31 - Prob. 89PP

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.

The amount of work required to bring an alpha particle (charge = + 2 e ) from rest at infinity to the “surface” of the nucleus.

Solution Summary: The author explains the amount of work required to bring an alpha particle from rest at infinity to the “surface” of the nucleus.

Concept explainers

The amount of work required to bring an alpha particle (charge =+2 e ) from rest at infinity to the “surface” of the nucleus.

Want to see the full answer?

Chapter 31 Solutions