/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 There are \(Z\) protons in the n... [FREE SOLUTION] | 91Ó°ÊÓ

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There are \(Z\) protons in the nucleus of an atom, where \(Z\) is the atomic number of the element. An \(\alpha\) particle carries a charge of \(+2 \mathrm{e}\). In a scattering experiment, an \(\alpha\) particle, heading directly toward a nucleus in a metal foil, will come to a halt when all the particle's kinetic energy is converted to electric potential energy. In such a situation, how close will an \(\alpha\) particle with a kinetic energy of \(5.0 \times 10^{-13} \mathrm{J}\) come to a gold nucleus \((Z=79) ?\)

Short Answer

Expert verified
The α particle will come as close as approximately 3.0 x 10^-14 meters to the gold nucleus.

Step by step solution

01

Understand the Scenario

An \( \alpha \) particle heads towards a gold nucleus. As it approaches, its kinetic energy is fully converted into electric potential energy where it momentarily comes to rest.
02

Identify the Variables

The kinetic energy \( K \) of the \( \alpha \) particle is given as \( 5.0 \times 10^{-13} \text{ J} \). The atomic number \( Z \) of gold is 79, and the charge on the \( \alpha \) particle is \(+2e\). The elementary charge \( e \) is \(1.6 \times 10^{-19} \text{ C} \).
03

Use the Conservation of Energy

At the closest point of approach, the kinetic energy (\( K \)) of the \( \alpha \) particle is converted into electric potential energy. Thus, \( K = U_e \), where \( U_e \) is the electric potential energy.
04

Calculate the Electric Potential Energy

The electric potential energy \( U_e \) between two point charges is given by \[ U_e = \frac{k \cdot q_1 \cdot q_2}{r} \]where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \), \( q_1 = +2e \), and \( q_2 = Ze \).
05

Solve for Distance \( r \)

Rearrange the formula to solve for \( r \):\[ r = \frac{k \cdot q_1 \cdot q_2}{K} \]Substitute \( q_1 = 2e \), \( q_2 = Ze \), and \( K = 5.0 \times 10^{-13} \text{ J} \):\[ r = \frac{(8.99 \times 10^9) (2 \times 1.6 \times 10^{-19}) (79 \times 1.6 \times 10^{-19})}{5.0 \times 10^{-13}} \].
06

Calculate the Result

Calculate \( r \):\[ r = \frac{8.99 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{5.0 \times 10^{-13}} \]\[ r \approx 3.0 \times 10^{-14} \text{ m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential Energy
Electric potential energy plays a crucial role in understanding how charged particles interact with each other. Imagine it like energy stored between two charges due to their positions. When you move these charges closer or further apart, you're dealing with electric potential energy.
In our context, an alpha particle is approaching a gold nucleus. The alpha particle has a positive charge, and so does the nucleus. As these positively charged bodies move closer, the electric potential energy between them increases. This energy becomes important as it is the point at which the alpha particle temporarily halts its motion.
The formula for electric potential energy between two point charges is: \[ U_e = \frac{k \cdot q_1 \cdot q_2}{r} \]where:
  • \( k \) is the Coulomb's constant,
  • \( q_1 \) and \( q_2 \) are the charges, and
  • \( r \) is the distance between the charges.
Consequently, as the alpha particle moves closer to the nucleus, the conversion of kinetic energy into electric potential energy halts its progress.
Kinetic Energy Conversion
Kinetic energy relates to the motion of an object. For the alpha particle coming toward a nucleus, its motion embodies its kinetic energy. But as it nears the nucleus, this energy undergoes conversion.
In the experiment, when the particle halts, it indicates that its kinetic energy has transformed completely into electric potential energy. This phenomenon follows the conservation of energy principle which states that energy cannot be created or destroyed, only converted from one form to another.
In practical terms, the kinetic energy \( K \) given to the alpha particle is \( 5.0 \times 10^{-13} \text{ J} \). Upon nearing the nucleus, the kinetic energy converts to electric potential energy at the pointing exactly where the particle stops its forward motion.
This conversion helps us determine how close the alpha particle can approach the nucleus before stopping completely.
Coulomb's Law
Coulomb's Law describes the interaction between two charged particles. It quantifies the amount of force between them based on their charges and the distance between them. The law is a cornerstone in understanding electrostatic forces.
The formula of Coulomb's Law is given as:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]Here, the symbols represent:
  • \( F \): the magnitude of force between the particles,
  • \( k \): Coulomb's constant, \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \),
  • \( q_1 \) and \( q_2 \): the electric charges, and
  • \( r \): the distance between the charges.
Coulomb's Law explains why the alpha particle stops. As it approaches the gold nucleus, the ever-increasing repulsive force due to their like charges brings the alpha particle to a standstill.
Atomic Number
The atomic number is a defining trait of an element, indicating how many protons reside in its nucleus. It's symbolized as \( Z \).
In our case, the gold nucleus has an atomic number of 79, denoting 79 protons. This information becomes critical in scaling the electric potential energy during the alpha particle scattering experiment.
Each proton carries a charge of \( e = 1.6 \times 10^{-19} \text{ C} \). Therefore, the charge of a gold nucleus is \( 79e \), a crucial value in calculating the forces and potential energy between the interacting particles.
Understanding the atomic number helps calculate the level of interaction between the charges, guiding us in solving how closely the alpha particle can approach the nucleus based on energy calculations.

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Most popular questions from this chapter

When an electron makes a transition between energy levels of an atom, there are no restrictions on the initial and final values of the principal quantum number \(n .\) According to quantum mechanics, however, there is a rule that restricts the initial and final values of the orbital quantum number \(\ell\). This rule is called a selection rule and states that \(\Delta \ell=\pm 1\). In other words, when an electron makes a transition between energy levels, the value of \(\ell\) can only increase or decrease by one. The value of \(\ell\) may not remain the same nor may it increase or decrease by more than one. According to this rule, which of the following energy level transitions are allowed? (a) \(2 \mathrm{s} \rightarrow 1 \mathrm{s}\) (b) \(2 p \rightarrow 1 s\) (c) \(4 p \rightarrow 2 p(d) 4 s \rightarrow 2 p\) (e) \(3 \mathrm{d} \rightarrow 3 \mathrm{s}\)

The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional \(2.86 \mathrm{eV}\) of energy. What is the quantum number \(n\) of the state into which the electron moves?

A sodium atom \((Z=11)\) contains 11 protons in its nucleus. Strictly speaking, the Bohr model does not apply, because the neutral atom contains 11 electrons instead of a single electron. However, we can apply the model to the outermost electron as an approximation, provided that we use an effective value \(Z_{\text {effective }}\) rather than 11 for the number of protons in the nucleus. (a) The ionization energy for the outermost electron in a sodium atom is \(5.1 \mathrm{eV}\). Use the Bohr model with \(Z=Z_{\text {effective }}\) to calculate a value for \(Z_{\text {effective }}\) (b) Using \(Z=11\) and \(Z=Z_{\text {effective }},\) determine the corresponding two values for the radius of the outermost Bohr orbit.

A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is \(514 \mathrm{nm}\) and the power is 1.5 W. During surgery, the laser beam is turned on for 0.050 s. During this time, how many photons are emitted by the laser?

Write down the fourteen sets of the four quantum numbers that correspond to the electrons in a completely filled \(4 \mathrm{f}\) subshell.

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