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A gold nucleus has a radius of \(7.3 \times 10^{-15} \mathrm{~m}\) and a charge of \(+79 e\). Through what voltage must an alpha particle, with charge \(+2 e,\) be accelerated so that it has just enough energy to reach a distance of \(2.0 \times 10^{-14} \mathrm{~m}\) from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

Short Answer

Expert verified
The alpha particle must be accelerated through a voltage of approximately \(5.2 \times 10^{6} V\) to reach a distance of \(2.0 \times 10^{-14}\) m from the surface of the gold nucleus.

Step by step solution

01

Understanding the provided information

Start by noting down the important details given in the problem. We know that the gold nucleus has a charge of \(+79 e\), where \(e\) is the unit charge of an electron. The radius of the gold nucleus is given as \(7.3 \times 10^{-15} \mathrm{~m}\). The alpha particle has a charge of \(+2 e\) and we need to find the voltage required to accelerate it to a distance of \(2.0 \times 10^{-14} \mathrm{~m}\) from the surface of the gold nucleus.
02

Calculate the total distance from the center

The distance of the alpha particle from the nucleus' surface is given, however, the calculation will require the total distance from the center of the nucleus. As such, add the radius of the gold nucleus to the specified distance to get this total distance. This will be \(7.3 \times 10^{-15} \mathrm{~m} + 2.0 \times 10^{-14} \mathrm{~m} = 2.73 \times 10^{-14} \mathrm{~m}\).
03

Application of the potential difference equation

Now, the potential difference a charge has to overcome to reach a certain distance from another charge can be calculated with the help of the following formula: \(V = k \frac{Q}{r}\), where \(V\) is the potential difference, \(k\) is Coulomb’s constant \(8.99 \times 10^{9} \mathrm{~m^2kg/s^2C^2}\), \(Q\) is the charge of the nucleus and \(r\) is the total distance. Plug in the values \(k = 8.99 \times 10^{9} \mathrm{~m^2kg/s^2C^2}, Q = +79e\) and \(r = 2.73 \times 10^{-14} m\), where \(e = 1.6 \times 10^{-19} C\) (the charge of an electron) in the formula in order to find the required voltage.
04

Final calculation

Carrying out the calculation, the voltage that an alpha particle must be accelerated through in order to just reach a distance of \(2.0 \times 10^{-14} \mathrm{~m}\) from the surface of the gold nucleus is \(V = 8.99 \times 10^{9} \frac{(79)(1.6 \times 10^{-19})}{2.73 \times 10^{-14}}\), which equals approximately \(5.2 \times 10^{6} V\).

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Key Concepts

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

Electric Potential Difference
When studying electricity, it's crucial to understand the electric potential difference concept, commonly referred to as voltage. Imagine voltage as the electrical pressure that propels charged particles through a circuit. It's analogous to water pressure driving water through pipes. Voltage is a measure of the work needed to move a charge from one point to another against an electric field.

In the context of our exercise, we seek the voltage required to accelerate an alpha particle towards a gold nucleus. The electric potential difference here is the energy per unit charge needed for the alpha particle to surmount the repulsive force of the nucleus's positive charge.

To calculate this, we look at how much work it would take to move the alpha particle to a certain distance from the nucleus. The higher the voltage, the more energy the alpha particle has at its disposal to overcome the electric repulsion exerted by the gold nucleus.
Coulomb's Law
Coulomb's law is a cornerstone of electrostatics, named after Charles-Augustin de Coulomb, who introduced the law. It quantifies the electric force between two charges. The law states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, Coulomb's law is expressed as: \[ F = k \frac{q_1 \cdot q_2}{r^2} \] where \(F\) is the force, \(q_1\) and \(q_2\) are the charges, \(r\) is the distance between the charges, and \(k\) is Coulomb's constant.

In our physics problem, the alpha particle is repelled by the gold nucleus because both carry a positive charge. By calculating the electric potential difference needed for the alpha particle to reach a certain point near the nucleus, we effectively apply Coulomb's law to determine the work done against this force of repulsion.
Nuclear Charge
The term 'nuclear charge' refers to the total charge within an atom's nucleus. It is the product of the number of protons in the nucleus and the elementary charge \(e\). Since protons are positively charged, the nuclear charge is always positive. Nuclear charge plays an integral role in the behavior of electrons in an atom, affecting their energy levels and the chemical properties of the element.

In the context of our problem, the nuclear charge of the gold nucleus is significant because it determines the strength of the electric field around the nucleus. The gold nucleus has 79 protons and thus a nuclear charge of \(+79e\). This large positive charge influences the potential difference required to accelerate the alpha particle. The higher the nuclear charge, the more potent the electric field and thus the greater the electric potential difference needed for an external charge to approach the nucleus.

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Most popular questions from this chapter

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil was sprayed in very fine drops (about \(10^{-4} \mathrm{~mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d\). A potential difference \(V_{A B}\) was maintained between the plates, causing a downward electric field between them. Some of the oil drops acquired a negative charge because of frictional effects or because of ionization of the surrounding air by x rays or radioactivity. The drops were observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates remained at rest if the magnitude of its charge was \(q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}\) where \(\rho\) is oil's density. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, Millikan determined the charge on that drop. provided its radius \(r\) was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determincd \(r\) by cutting off the electric field and measuring the terminal speed \(u_{\mathrm{i}}\) of the drop as it fell. (We discussed terminal speed in Section \(\left.5.3 .\right)\) The viscous force \(F\) on a sphere of radius \(r\) moving at speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When a drop fell at \(v_{1}\), the viscous force just balanced the drop's weight \(w=m g\). Show that the magnitude of the charge on the drop was \(q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{1}^{3}}{2 \rho g}}\) (c) You repeat the Millikan oil-drop experiment. Four of your measured values of \(V_{A B}\) and \(v_{1}\) are listed in the table: $$ \begin{array}{lcccc} \text { Drop } & 1 & 2 & 3 & 4 \\ \hline V_{A B}(\mathrm{~V}) & 9.16 & 4.57 & 12.32 & 6.28 \\ v_{1}\left(10^{-5} \mathrm{~m} / \mathrm{s}\right) & 2.54 & 0.767 & 4.39 & 1.52 \end{array} $$ In your apparatus, the separation \(d\) between the horizontal plates is \(1.00 \mathrm{~mm}\). The density of the oil you use is \(824 \mathrm{~kg} / \mathrm{m}^{3}\). For the viscosity \(\eta\) of air, use the value \(1.81 \times 10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). Calculate the charge \(q\) of each drop. (d) If electric charge is quantized (that is, exists in multiples of the magnitude of the charge of an electron), then the charge on each drop is -ne, where \(n\) is the number of excess electrons on each drop. (All four drops in your table have negative charge.) Drop 2 has the smallest magnitude of charge observed in the experiment, for all 300 drops on which measurements were made, so assume that its charge is due to an excess charge of one electron. Determine the number of excess electrons \(n\) for each of the other three drops. (e) Use \(q=-n e\) to calculate \(e\) from the data for each of the four drops, and average these four values to get your best experimental value of \(e .\)

In a certain region of space the electric potential is given by \(V=+A x^{2} y-B x y^{2},\) where \(A=5.00 \mathrm{~V} / \mathrm{m}^{3}\) and \(B=8.00 \mathrm{~V} / \mathrm{m}^{3}\) Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x=2.00 \mathrm{~m}, y=0.400 \mathrm{~m},\) and \(z=0\)

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

A particle with charge \(+7.60 \mathrm{nC}\) is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved \(8.00 \mathrm{~cm}\), the additional force has done \(6.50 \times 10^{-5} \mathrm{~J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{~J}\) of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(1.67 \times 10^{-27} \mathrm{~kg},\) charge \(+1.60 \times 10^{-19} \mathrm{C}\) ) is \(18.0 \mathrm{~cm}\) from the line and moving directly toward the line at \(3.50 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

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