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Chapter 29: Problem 7

How many protons are found in a \(^{136}\) Xe nucleus?

Short Answer

Expert verified
Answer: There are 54 protons in a \(^{136}\) Xe nucleus.

Step by step solution

01

Identify the atomic number of Xe

Check a periodic table to find the atomic number of Xenon (Xe). The atomic number represents the number of protons in the nucleus. In the case of Xenon, the atomic number is 54.
02

Protons in \(^{136}\) Xe nucleus

Since the atomic number of Xenon is 54, there are 54 protons in the nucleus of a \(^{136}\) Xe atom.

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Most popular questions from this chapter

An \(\alpha\) particle with a kinetic energy of \(1.0 \mathrm{MeV}\) is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the \(\alpha\) particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the \(\alpha\) particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"? (c) What is the minimum initial kinetic energy of an \(\alpha\) particle that will make contact with the gold nucleus?
A space rock contains \(3.00 \mathrm{~g}\) of \({ }_{62}^{147} \mathrm{Sm}\) and \(0.150 \mathrm{~g}\) of ${ }_{60}^{143} \mathrm{Nd} .{ }_{62}^{147} \mathrm{Sm} \alpha\( decays to \){ }_{60}^{143} \mathrm{Nd}\( with a half-life of \)1.06 \times 10^{11} \mathrm{yr}\(. If the rock originally contained no \){ }_{60}^{143} \mathrm{Nd}$ how old is it?
What is the binding energy of an \(\alpha\) particle (a \(^{4} \mathrm{He}\) nucleus)? The mass of an \(\alpha\) particle is \(4.00151 \mathrm{u}\).
\(_{52}^{106} \mathrm{Te}\) is radioactive; it \(\alpha\) decays to $^{102} \mathrm{s}_{0} \mathrm{Sn} .\( " \)_{50}^{102} \mathrm{Sn}$ is itself radioactive and has a half-life of 4.5 s. At \(t=0,\) a sample contains 4.00 mol of \(^{106}\) Te and 1.50 mol of \(^{102}\) so Sn. At \(t=\) \(25 \mu \mathrm{s},\) the sample contains \(3.00 \mathrm{mol}\) of \(^{106} \mathrm{Te}\) and $2.50 \mathrm{mol}\( of \)^{102} \mathrm{s} 0$ Sn. How much 102 50 Sn will there be at \(t=50 \mu \mathrm{s} ?\)
The radioactive decay of \(^{238} \mathrm{U}\) produces \(\alpha\) particles with a kinetic energy of \(4.17 \mathrm{MeV} .\) (a) At what speed do these \(\alpha\) particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that \(\alpha\) particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the \(\alpha\) particles using a velocity selector. If your magnet produces a magnetic field of \(0.30 \mathrm{T},\) what strength electric field would allow the \(\alpha\) particles to pass through undeflected? (c) Now that you know the speed of the \(\alpha\) particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the \(\alpha\) particle, what would the radius be in a \(0.30-\mathrm{T}\) field? (d) Why can you determine only the charge-to-mass ratio \((q / m)\) by this experiment, but not the individual values of \(q\) and \(m ?\)
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