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A radioactive source in the form of a metal sphere of diameter \(3.2 \times 10^{-3} m\) emits \(\beta\)-particle at a constant rate of \(6.25 \times 10^{10}\) particle/sec. The source is electrically insulated and all the \(\beta\)-particle are emitted from the surface. The potential of the sphere will rise to \(1 V\) in time. Find the time in which potential of the sphere becomes \(1 V\).

Short Answer

Expert verified
Time needed is approximately \( 1.78 \times 10^{-5} \, \text{s} \).

Step by step solution

01

Understanding the Problem

We need to determine the time required for the emitted \( \beta \)-particles to cause the potential of the sphere to increase to 1 V. We'll first identify the key parameters, including the charge of the \( \beta \)-particles and the relationship between potential, charge, and time.
02

Calculating the Charge Required for a Potential of 1 V

The formula for the potential \( V \) of a sphere is \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant (\( 8.99 \times 10^{9} \, \text{N m}^2/\text{C}^2 \)), \( Q \) is the charge, and \( r \) is the radius of the sphere. We set \( V = 1 \, \text{V} \) and the radius \( r \) as half the diameter, \( 1.6 \times 10^{-3} \, \text{m} \). Solve for \( Q \): \[ Q = \frac{Vr}{k} = \frac{1 \, \text{V} \times 1.6 \times 10^{-3} \, \text{m}}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \approx 1.78 \times 10^{-13} \, \text{C} \].
03

Determine Charge Accumulation Rate

Since each \( \beta \)-particle has a charge equivalent to that of an electron, we use the charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \). The sphere emits \( \beta \)-particles at \( 6.25 \times 10^{10} \) particles/sec, so the charge per second is: \[ \text{Charge per second} = 6.25 \times 10^{10} \times 1.6 \times 10^{-19} = 1 \times 10^{-8} \, \text{C/s} \].
04

Calculate Time for Potential to Rise to 1 V

To find the time \( t \), we use the formula \( Q = It \), where \( I \) is the current (charge per second). Rearrange for \( t \): \[ t = \frac{Q}{I} = \frac{1.78 \times 10^{-13} \, \text{C}}{1 \times 10^{-8} \, \text{C/s}} \approx 1.78 \times 10^{-5} \, \text{s} \].

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Key Concepts

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

Coulomb's law
Coulomb's law is a fundamental principle that describes how electric charges interact. It states that the force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, Coulomb's law can be expressed as:
  • \[ F = k \frac{{|q_1 q_2|}}{{r^2}} \]
Here, \( F \) represents the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, \( r \) is the distance between the centers of the two charges, and \( k \) is Coulomb's constant (\(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\)). This law helps us understand how electric forces can either pull charges together or push them apart.
Understanding Coulomb's law is essential in calculating the potential of charged objects, as it helps to determine the electric field around them. This is particularly important in scenarios involving insulators and conductors where charge distribution impacts electric potential.
electric potential
Electric potential, often referred to as voltage, is a measure of the potential energy per unit charge at a point in an electric field. It tells us how much potential energy a charge would have if placed at that point. The potential energy is influenced by both the amount of charge and the configuration of the electric field.
The electric potential \( V \) at a point due to a point charge \( Q \) is given by the formula:
  • \[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point of interest.
In the context of a sphere, as mentioned in the problem statement, this formula explains how the potential changes based on the accumulated charge and the size of the sphere (determined by the radius). When the sphere releases \(\beta\)-particles, each adding charge equivalent to an electron, the potential of the sphere increases, impacting the time required for the potential to reach a certain level, such as 1 V. This concept is crucial for understanding how devices store and utilize energy through electric charges.
charge of an electron
The charge of an electron is one of the fundamental constants in nature, crucial to understanding atomic and subatomic interactions. It is denoted as \( e \) and has a value of approximately \( 1.6 \times 10^{-19} \, \text{C} \) (Coulombs).
  • Electrons are negatively charged particles, and their charge is equal and opposite to that of protons.
  • This tiny charge value underlies the interactions in most electrical phenomena we experience, including the operation of electronics and chemical bonding.
In the exercise, knowing the charge of an electron allows us to calculate the rate of charge accumulation on a surface when a certain number of \( \beta\)-particles, which have the same charge as electrons, are emitted per second.
This information is then used to determine how long it takes for the potential of an object, such as a metal sphere, to reach a specified level. Understanding the charge of an electron is therefore pivotal for understanding how charges build up in various systems.

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Most popular questions from this chapter

The count rate from \(100 \mathrm{~cm}^{3}\) of a radioactive liquid is ' \(c\) '. Some of this liquid is now removed. The count rate of the remaining liquid is found to be \(c / 10\) after three half lives. Find the volume of the remaining liquid (in \(\mathrm{cm}^{3}\) )

Consider the nuclear reaction $$ { }_{14}^{28} \mathrm{Si}+\gamma \rightarrow{ }_{12}^{24} \mathrm{Mg}+\mathrm{X} $$ where \(\mathrm{X}\) is a nuclide. (a) What are \(Z\) and \(A\) for the nuclide \(X ?\) (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \({ }_{14}^{28} \mathrm{Si}\) atom is \(27.976927 \mathrm{u}\), and the mass of \(\mathrm{a}_{12}^{24} \mathrm{Mg}\) atom is \(23.985042 \mathrm{u}\).

A bone fragment found in a cave believed to have been inhabited by early humans contains \(0.29\) times as much \({ }^{14} \mathrm{C}\) as an equal amount of carbon in the atmosphere when the organism containing the bone died. (See Example \(19.4\) in Section 19.4.) Find the approximate age of the fragment. \((\ln 0.29=-1.209)\)

What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \({ }_{94}^{239} \mathrm{Pu} ;\) (b) \(\beta^{-}\)decay of \(_{11}^{24} \mathrm{Na} ;\) (c) \(\beta^{+}\)decay of \({ }_{8}^{15} \mathrm{O}\).

A thin uniform radioactive coating of an \(a\)-emitter is applied on a metallic sphere of radius ' \(a\) '. The sphere is connected to the ground by means of a thin wire of resistance ' \(R\) ' as shown in figure. Assume that, the sphere is uncharged at \(t=0\), the initial activity of \(a\) - emitter is \(A_{0}\) and the disintegration constant of radioactive material is \(\lambda=3 \sec ^{-1}\). Also assume that all the a-particles emitted go out of the sphere. Find the time (in sec) at which charge on the sphere is maximum. Use \(\ln 2=0.69\) and \(\frac{1}{4 \pi \varepsilon_{0} a}=2 R \lambda\).

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