A medical lab has 16gram of sample of radioactive isotope after 6hours it found that 12gm of sample have decayed the half life of isotope is?

Question

Nathan Ashcroft · Accepted Answer

#"3 hour"#

General equation for radioactivity is

#N = N_0 e^(-λt) color(white)(...)....[1]#

Where

Amount of radioactive isotope left after 6 hours is, 16 g - 12 g = 4 g

#"4 g" = "16 g" × e^(-λ * "6 hr")#

#(4 cancel"g")/(16 cancel"g") = e^(-λ * "6 hr")#

#1/4 = e^(-λ * "6 hr")#

#4 = e^(λ * "6 hr")#

Apply natural log on both the sides

#ln4= "6 hr" × λ#

#λ = ln4 / "6 hr"#

#ln2/"T"_"1/2" = ln4/"6 hr" color(white)(..)[∵ λ = ln2/"T"_"1/2"]#

#cancel(ln2)/"T"_"1/2" = (2 cancel(ln2))/"6 hr" color(white)(..)[∵ lnx^n = nlnx]#

#"T"_(1//2) = "6 hr" / 2 = color(blue)"3 hr"#

ALTERNATIVE APPROACH

Equation #[1]# can also be written as

#N = N_0/(2^"n") color(white)(...)....[2]#

Where

#"4 g" = "16 g"/2^"n"#

#2^"n" = "16 g"/"4 g"#

#2^"n" = 4#

#2^"n" = 2^2#

#"n" = 2#

Since we got #"n = 2"#, we can say that 6 hours is 2 half lives.

∴ Half life #= "6 hr"/2 = color(blue)"3 hr"#

Lily Adams · Answer

undefined The half-life of the radioactive isotope can be calculated using the formula for exponential decay:

$$ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{\frac{1}{2}}}} $$

Where:
- $ N(t) $ is the amount of the radioactive isotope remaining after time $ t $,
- $ N_0 $ is the initial amount of the radioactive isotope,
- $ T_{\frac{1}{2}} $ is the half-life of the radioactive isotope.

Given:
- $ N(t) = 12 $ grams (amount decayed after 6 hours),
- $ N_0 = 16 $ grams (initial amount),
- $ t = 6 $ hours.

Plug the given values into the formula and solve for $ T_{\frac{1}{2}} $:

$$ 12 = 16 \times \left(\frac{1}{2}\right)^{\frac{6}{T_{\frac{1}{2}}}} $$

$$ \frac{12}{16} = \left(\frac{1}{2}\right)^{\frac{6}{T_{\frac{1}{2}}}} $$

$$ \frac{3}{4} = \left(\frac{1}{2}\right)^{\frac{6}{T_{\frac{1}{2}}}} $$

$$ \frac{3}{4} = \left(\frac{1}{2}\right)^{\frac{6}{T_{\frac{1}{2}}}} = \left(2^{-1}\right)^{\frac{6}{T_{\frac{1}{2}}}} = 2^{-\frac{6}{T_{\frac{1}{2}}}} $$

$$ 2^{\frac{6}{T_{\frac{1}{2}}}} = \frac{4}{3} $$

$$ \frac{6}{T_{\frac{1}{2}}} = \log_2\left(\frac{4}{3}\right) $$

$$ T_{\frac{1}{2}} = \frac{6}{\log_2\left(\frac{4}{3}\right)} $$

Calculate the value of $ T_{\frac{1}{2}} $ to find the half-life of the isotope.