/*! 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 92 A sample of an alpha emitter hav... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of an alpha emitter having an activity of \(0.18 \mathrm{Ci}\) is stored in a \(25.0\) - \(\mathrm{mL}\) sealed container at \(22{ }^{\circ} \mathrm{C}\) for 245 days. (a) How many alpha particles are formed during this time? (b) Assuming that each alpha particle is converted to a helium atom, what is the partial pressure of helium gas in the container after this 245 -day period?

Short Answer

Expert verified
After 245 days, approximately \(1.409 \times 10^{17}\) alpha particles are formed, which are converted into helium atoms. The partial pressure of helium gas in the container becomes approximately \(2.33 \times 10^{-4}\) atm.

Step by step solution

01

Calculation of the number of alpha particles formed

First, we need to convert the activity of the alpha emitter from curies (Ci) to decays per second (Bq). We know that 1 Ci = 3.7 × 10^10 Bq. So, the activity of the sample in Bq is: Activity (in Bq) = 0.18 Ci × (3.7 × 10^10 Bq/Ci) = 6.66 × 10^9 Bq Now, let's calculate the total number of decays (alpha particles) formed during 245 days: Number of alpha particles formed = Activity (in Bq) × Time (in seconds) First, convert 245 days to seconds: 245 days × (24 hours/day) × (3600 seconds/hour) = 2.116 × 10^7 seconds And now, calculate the number of alpha particles formed: Number of alpha particles formed = 6.66 × 10^9 Bq × 2.116 × 10^7 seconds ≈ 1.409 × 10^17 alpha particles
02

Calculation of the number of moles of helium gas formed

As mentioned in the problem, we assume that each alpha particle is converted to a helium atom. Therefore, the number of helium atoms formed is equal to the number of alpha particles formed which is 1.409 × 10^17. To calculate the number of moles, we will use Avogadro's number (6.022 × 10^23 atoms/mole): Number of moles of helium gas (n) = (1.409 × 10^17 helium atoms) / (6.022 × 10^23 atoms/mole) ≈ 2.34 × 10^-7 moles
03

Calculation of the partial pressure of helium gas in the container

Now we will use the Ideal Gas Law to find the partial pressure of helium gas in the container: PV = nRT where P is the partial pressure (in atm), V is the volume (in liters), n is the number of moles of helium gas, R is the gas constant (0.0821 L atm/mol K), and T is the temperature (in Kelvin). First, convert 25.0 mL to liters and 22°C to Kelvin: V = 25.0 mL × (1 L / 1000 mL) = 0.025 L T = 22 + 273.15 = 295.15 K Now, using the Ideal Gas Law, we can find the partial pressure of helium gas: P = nRT / V P = (2.34 × 10^-7 moles × 0.0821 L atm/mol K × 295.15 K) / 0.025 L ≈ 2.33 × 10^-4 atm So, after 245 days, the partial pressure of helium gas in the container is approximately \(2.33 \times 10^{-4}\) atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Radioactive Decay
Radioactive decay is a fundamental concept in nuclear physics and chemistry, describing the process by which an unstable atomic nucleus loses energy by emitting radiation. Typically, this involves the transformation of one element into another and occurs naturally in all radioactive substances.

Alpha decay is a type of radioactive decay where an alpha particle (consisting of two protons and two neutrons) is emitted from the nucleus, effectively reducing the original element's atomic number by two and its mass number by four. The decay rate, or activity, is often measured in units like curies (Ci) or becquerels (Bq), with one curie being equivalent to 3.7 x 10^10 decays per second.

For students attempting to grasp the calculations involved, it's essential to understand that the decay rate represents how many alpha particles are emitted per time unit. By multiplying this rate by the elapsed time, we can determine the total number of alpha particles produced over a given period, as demonstrated in the exercise's solution.
Applying the Ideal Gas Law
The Ideal Gas Law is a crucial equation in chemistry that relates the pressure (P), volume (V), temperature (T), and amount of gas (n) in moles. Expressed as the equation PV = nRT, the Ideal Gas Law allows us to predict the behavior of an ideal gas under different conditions. In this formula, 'R' stands for the universal gas constant, which is approximately 0.0821 L atm/mol K.

In the context of our exercise, once the number of alpha particles—now helium atoms—has been determined, we can use the Ideal Gas Law to calculate the partial pressure of helium within the container. We assume that our gas behaves ideally, an assumption that simplifies real-world behavior. To perform this calculation correctly, it's necessary to ensure that all the values are in the appropriate units: volume in liters, temperature in Kelvin, and pressure in atmospheres.

A common point of confusion can be unit conversion—for instance, converting milliliters to liters or Celsius to Kelvin, which is vital for accurate calculations as seen in the step-wise solution.
The Significance of Avogadro's Number
Avogadro's number, approximately 6.022 x 10^23, is an incredibly large constant that denotes the number of particles, often atoms or molecules, in one mole of a substance. This number is pivotal in translating from the microscopic world of atoms and molecules to the macroscopic world of grams and liters that we can measure in a laboratory.

When dealing with radioactive decay and the production of gas, understanding Avogadro's number is fundamental as it enables the conversion from the number of individual atoms or molecules produced to moles. This is precisely what the exercise demonstrates: transitioning from the quantity of helium atoms to moles. By doing so, we use Avogadro's number as a bridge, allowing us to utilize the Ideal Gas Law and determine properties like pressure.

When tackling problems like these, remember that the clarity of the unit conversions and the direct application of constants such as Avogadro's number will help ensure the accuracy of your results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wooden artifact from a Chinese temple has a \({ }^{14} \mathrm{C}\) activity of \(38.0\) counts per minute as compared with an activity of \(58.2\) counts per minute for a standard of zero age. From the halflife for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}\), determine the age of the artifact.

Putting Concepts Together Potassium ion is present in foods and is an essential nutrient in the human body. One of the naturally occurring isotopes of potassium, potassium- 40 , is radioactive. Potassium-40 has a natural abundance of \(0.0117 \%\) and a half- life \(t_{1 / 2}=1.28 \times 10^{9} \mathrm{yr}\). It undergoes radioactive decay in three ways: \(98.2 \%\) is by electron capture, \(1.35 \%\) is by beta emission, and \(0.49 \%\) is by positron emission. (a) Why should we expect \({ }^{40} \mathrm{~K}\) to be radioactive? (b) Write the nuclear equations for the three modes by which \({ }^{40} \mathrm{~K}\) decays. (c) How many \({ }^{40} \mathrm{~K}^{+}\)ions are present in \(1.00 \mathrm{~g}\) of \(\mathrm{KCl}\) ? (d) How long does it take for \(1.00 \%\) of the \({ }^{40} \mathrm{~K}\) in a sample to undergo radioactive decay? SOLUTION (a) The \({ }^{40} \mathrm{~K}\) nucleus contains 19 protons and 21 neutrons. There are very few stable nuclei with odd numbers of both protons and neutrons (Section 21.2). (b) Electron capture is capture of an inner-shell electron by the nucleus: $$ { }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar} $$ Beta emission is loss of a beta particle \((-1 \mathrm{e})\) ) by the nucleus: $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e} $$ Positron emission is loss of a positron \(\left(+{ }_{+}^{0} \mathrm{e}\right)\) by the nucleus: $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar}+{ }_{+1}^{0} \mathrm{e} $$ (c) The total number of \(\mathrm{K}^{+}\)ions in the sample is $$ (1.00 \mathrm{~g} \mathrm{KCl})\left(\frac{1 \mathrm{~mol} \mathrm{KCl}}{74.55 \mathrm{~g} \mathrm{KCl}}\right)\left(\frac{1 \mathrm{~mol} \mathrm{~K}}{1 \mathrm{~mol} \mathrm{KCl}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{~K}^{+}}{1 \mathrm{~mol} \mathrm{~K}^{+}}\right)=8.08 \times 10^{21} \mathrm{~K}^{+} \text {ions } $$ Of these, \(0.0117 \%\) are \({ }^{40} \mathrm{~K}^{+}\)ions: $$ \left(8.08 \times 10^{21} \mathrm{~K}^{+} \text {ions }\right)\left(\frac{0.0117^{40} \mathrm{~K}^{+} \text {ions }}{100^{+} \text {ions }}\right)=9.45 \times 10^{17} \text { potassium-40 ions } $$ (d) The decay constant (the rate constant) for the radioactive decay can be calculated from the half-life, using Equation 21.20: $$ k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{1.28 \times 10^{9} \mathrm{yr}}=\left(5.41 \times 10^{-10}\right) / \mathrm{yr} $$ The rate equation, Equation \(21.19\), then allows us to calculate the time required: $$ \begin{aligned} \ln \frac{N_{t}}{N_{0}} &=-k t \\ \ln \frac{99}{100} &=-\left[\left(5.41 \times 10^{-10}\right) / \mathrm{yr}\right] t \\ -0.01005 &=-\left[\left(5.41 \times 10^{-10}\right) / \mathrm{yr}\right] t \\ t &=\frac{-0.01005}{\left(-5.41 \times 10^{-10}\right) / \mathrm{yr}}=1.86 \times 10^{7} \mathrm{yr} \end{aligned} $$ That is, it would take \(18.6\) million years for just \(1.00 \%\) of the \({ }^{40} \mathrm{~K}\) in a sample to decay.

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min}\). (a) How much loss of mass from the Sun occurs in one day from just the encrgy falling on Farth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{amu} ;{ }^{141} \mathrm{Ba}\) nuclear mass, \(140.8833 \mathrm{amu} ;{ }^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in \(1.0\) day?

Indicate the number of protons and neutrons in the following nuclei: (a) \({ }_{53}^{129} \mathrm{I}\), (b) \({ }^{138} \mathrm{Ba}\), (c) neptunium-237.

Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits \(8 \times 10^{-13} \mathrm{~J}\) of energy and if the average person weighs \(75 \mathrm{~kg}\), calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.