Problem 1 - Applied Physics

Question

Find the mass in kilograms of the \({ }_{92}^{232} \mathrm{U}\) atom if its mass in atomic mass units is \(232.037131 \mathrm{u}\).

Step-by-Step Solution

Answer

The mass of \( _{92}^{232} \mathrm{U} \) is \( 3.8551335 \times 10^{-25} \ kg \).

Step 1: Understand Atomic Mass Unit (amu)

1 atomic mass unit (amu), denoted by "u", is defined as one twelfth of the mass of a carbon-12 atom. It is equivalent to \( 1.66053906660 \times 10^{-27} \ kg \). The given mass of \( _{92}^{232} \mathrm{U} \) in atomic mass units is 232.037131 u.

Step 2: Calculate Mass in Kilograms

To convert the given mass from atomic mass units to kilograms, multiply it by the conversion factor \( 1.66053906660 \times 10^{-27} kg/u \). \[\text{Mass in kilograms} = 232.037131 \, \text{u} \times 1.66053906660 \times 10^{-27} \, \text{kg/u} = 3.855133548 \times 10^{-25} \, kg\]

Step 3: Round the Answer

Express the product from the previous step to an appropriate number of significant figures based on the original data provided.In this case, the mass of \( _{92}^{232} \mathrm{U} \) given as 232.037131 u has eight significant figures, so we should express the result in eight significant figures: \[3.8551335 \times 10^{-25} kg \].

Key Concepts

Atomic Mass Conversion

Atomic mass conversion is a critical step when dealing with the mass of particles at the atomic scale. Atoms and subatomic particles are so tiny that using a traditional measurement like the kilogram is impractical. Instead, we use the atomic mass unit (amu) as a standard to denote their mass. An amu is precisely defined as one-twelfth of the mass of a carbon-12 atom. This small scale measurement equates to a tiny quantity in kilogram terms—precisely, each amu equals \(1.66053906660 \times 10^{-27}\) kg. When converting from atomic mass units to kilograms, which is often necessary in scientific calculations, we need to multiply the atomic mass by this conversion factor. For instance, if you have a uranium-232 atom with a mass of 232.037131 amu, you convert it by multiplying it by \(1.66053906660 \times 10^{-27}\) kg/amu. This process will give you the mass in kilograms, which is crucial for further scientific calculations and understanding atomic interactions on a more relatable scale.

Mass Calculation

Calculating mass involves mathematical conversions, especially in scientific settings. When given the mass of an element or atom in atomic mass units, the task often requires converting that mass into kilograms for standardized scientific examination and comparison.The basic formula for mass calculation in this context is:\[\text{Mass in kilograms} = \text{Mass in atomic mass units} \times \text{Conversion factor} \]For example, for a uranium-232 atom, we are given 232.037131 amu. By applying the conversion factor (\(1.66053906660 \times 10^{-27}\) kg/u), you'll get:\[232.037131 \times 1.66053906660 \times 10^{-27} = 3.855133548 \times 10^{-25} \text{ kg}\]It's important to ensure the products of your calculations are accurate and relevant to the problem requirement. Scientific notations are often used as they make it easier to handle extremely large or small numbers. It also streamlines the readability and reduces chances of error when dealing with numerous zeros.

Significant Figures

In scientific calculations, the accuracy of your measurements and conversion results is paramount. This is where significant figures come into play. They represent the precision of your data and ensure the reliability of the reported results.Significant figures include all known digits in a number, plus one estimated digit. In our example with the uranium atom, we start with 232.037131 amu, which has eight significant figures. Thus, any computed result must maintain this level of precision, ensuring no loss of detail or introduction of error due to rounding.When successful, calculations should showcase the same amount of significant figures. For our uranium-232 mass in kilograms, we achieve this by expressing the final result as:\[3.8551335 \times 10^{-25} \text{ kg}\]By rounding to eight significant figures—not simply eight decimal places—we ensure that the precision of the value is intact. Especially when small variances significantly impact scientific interpretations and conclusions, mindful use of significant figures is crucial to communicating accuracy and maintaining scientific integrity.