The diameter of the atomic nucleus of a lead atom is 1.75*10^15 m.
I need help in making a word problem out of this given data I am confused at what to do or what equation to start with
To create a word problem from the given data, you can focus on understanding the relationship between the diameter of an atomic nucleus and the size of a lead atom. Here's an example:
Word Problem: "Consider a lead atom with an atomic nucleus. The diameter of the atomic nucleus is measured to be 1.75 x 10^15 meters. Imagine that you are a scientist studying atomic structures, and you want to compare the size of the atomic nucleus to everyday objects. How can you illustrate this size difference?
The average diameter of a human hair is approximately 100 micrometers. If we were to scale down the diameter of the atomic nucleus to human hair size, how many times smaller would it be?
Hint: Remember that 1 meter is equal to 10^6 micrometers."
By introducing the concept of scaling down the diameter of the atomic nucleus to the size of a human hair, you can create a word problem that engages readers and encourages them to think about the vast difference in size between atomic particles and everyday objects.
To create a word problem using the given data, you can focus on finding other quantities or using relationships related to atomic nuclei and lead atoms. Here's an example word problem you can create:
"Lead is a heavy element commonly used in radiation shielding. Imagine a scenario where a scientist needs to design a lead box to contain a small radioactive sample. The scientist knows that the diameter of a lead atom's nucleus is 1.75*10^15 m.
If the scientist wants to ensure that the sample remains isolated and properly shielded, what should be the minimum thickness of the lead box material in order to prevent any radiation from escaping? Assume that radiation can't pass through the lead material.
Hint: Use the diameter of the lead atom's nucleus to find the minimum thickness of the lead box material."
To solve this word problem, you will need to use the given diameter of the atomic nucleus and apply some scientific principles. Here's a possible approach:
1. Recognize that the diameter of the atomic nucleus represents the maximum distance a radiation particle might travel from the nucleus, as it is the size at which the "edge" of the nucleus is located.
2. Since we want to prevent any radiation from escaping, the minimum thickness of the lead box material should be equal to or greater than the maximum distance a radiation particle might travel from the nucleus.
3. Recall that the radius of a sphere is half of its diameter. Therefore, the radius (r) of the atomic nucleus is (1.75*10^15 m)/2 = 8.75*10^14 m.
4. To prevent radiation from escaping, the minimum thickness of the lead box material should be equal to or greater than twice the radius of the atomic nucleus (2r). Calculate this value by multiplying the radius by 2.
5. The minimum thickness of the lead box material, 2r, is equal to 2*(8.75*10^14 m) = 1.75*10^15 m.
Therefore, the minimum thickness of the lead box material should be 1.75*10^15 m in order to prevent any radiation from escaping.
You can further customize the word problem by introducing other variables or adding more context based on your specific needs.