Question #1dc8f

May 1, 2016

It is four.

Explanation:

The half life of any substance is given by:
${t}_{\frac{1}{2}} = \frac{I n 2}{l} a m \mathrm{da}$

Hence

${t}_{\frac{1}{2} \left(1\right)} = \frac{I n 2}{l} a m {\mathrm{da}}_{1}$ .... and.... ${t}_{\frac{1}{2} \left(2\right)} = \frac{I n 2}{l} a m {\mathrm{da}}_{2}$

We want to find:

$R a t i o = {t}_{\frac{1}{2} \left(1\right)} / \left({t}_{\frac{1}{2} \left(2\right)}\right)$

Substituting the previous equations into the above gives
$R a t i o = {t}_{\frac{1}{2} \left(1\right)} / \left({t}_{\frac{1}{2} \left(2\right)}\right) = l a m {\mathrm{da}}_{2} / l a m {\mathrm{da}}_{1}$ [equation A]

For a sample of N atoms, the activity, A, is given by:

$A = l a m \mathrm{da} \cdot N$

Where $l a m \mathrm{da}$ is the decay constant.

So for sample 1: ${A}_{1} = l a m {\mathrm{da}}_{1} \cdot {N}_{1}$ [equation 1]

and for sample 2: ${A}_{2} = l a m {\mathrm{da}}_{2} \cdot {N}_{2}$ [equation 2]

We are told that that ${N}_{1} = 2 {N}_{2}$ or${N}_{1} / 2 = {N}_{2}$
and also ${A}_{2} = 2 {A}_{1}$

Substituting both of these into equation 2 we get:
$2 {A}_{1} = l a m {\mathrm{da}}_{2} \cdot {N}_{1} / 2$
So
${A}_{1} = l a m {\mathrm{da}}_{2} / 4 \cdot {N}_{1}$ [equation 3]

Comparing equation 1 and 3 we can see that:

$l a m {\mathrm{da}}_{1} = l a m {\mathrm{da}}_{2} / 4$

and so $l a m {\mathrm{da}}_{2} / l a m {\mathrm{da}}_{1} = 4$

Hence the ratio of halve lives is 4 (by comparing this result to equation A)
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/5725a03d7c014928e2d1dc8f", "fetch_time": 1571248098000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00311.warc.gz", "warc_record_offset": 702827196, "warc_record_length": 6276, "token_count": 620, "char_count": 1483, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "longest", "language": "en", "language_score": 0.6880872249603271, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}