Add Numbers in Scientific Notation Quick Check

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Question
The half-life of uranium-235 is 700,000,000 years. The half-life of uranium-238 is 4.463×1011
years longer. What is the half-life of uranium-238? The answer should be in decimal form.(1 point)
Responses

447,000,000,000
447,000,000,000

11,463,000,000
11,463,000,000

4.4637×1011
4.4637 times 10 superscript 11 baseline

516,300,000,000

1 answer

To find the half-life of uranium-238, you first need to calculate the half-life of uranium-238 based on the information given:

The half-life of uranium-235 is:

\[ 7.0 \times 10^{8} , \text{years} \] (700,000,000 years).

The half-life of uranium-238 is given as:

\[ 4.463 \times 10^{11} , \text{years} \text{ longer} \]

To find the half-life of uranium-238, you add the half-life of uranium-235 to the additional years:

\[ \text{Half-life of uranium-238} = (4.463 \times 10^{11}) + (7.0 \times 10^{8}) \]

To perform this operation, convert \( 7.0 \times 10^{8} \) to a compatible form:

\[ 7.0 \times 10^{8} = 0.0007 \times 10^{11} \]

Now you can add the two:

\[ (4.463 \times 10^{11}) + (0.0007 \times 10^{11}) = (4.4637 \times 10^{11}) \]

Now converting \( 4.4637 \times 10^{11} \) back to decimal form gives:

\[ 4.4637 \times 10^{11} = 446,370,000,000 \]

However, when looking at the options provided, we don't see this answer. Since it seems like the operation is correct, we can look at the original numbers again in standard decimal format.

Calculating:

  1. \( 4.463 \times 10^{11} = 446,300,000,000 \)
  2. Adding \( 700,000,000 \) (or \( 0.7 \times 10^{9} = 0.0007 \times 10^{11} \))

The correct half-life of uranium-238 becomes:

\[ 4.4637 \times 10^{11} \text{ years}, \]

This can also be presented as:

\[ 446,370,000,000 \text{ years} \]

Given the closest options, 447,000,000,000 would be the best choice as it's the correct order of magnitude in context, and it matches our calculated result closely.

So the answer to the question is:

447,000,000,000 years.

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