# If the half life of radium D is 22 years, then it will take __ years to decrease to 10%.

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If the half life of radium D is 22 years, then it will take __ years to decrease to 10%.
1. 29
2. 39
3. 73.11
4. 90

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Correct Answer - Option 3 : 73.11

Concept:

• Radioactivity: The atoms having a number of neutrons much more than proton are unstable. The nuclei of such atoms exhibit radioactivity.
• An example of Such an Atom is U - 238, where the number of Neutrons is 146, and the number of protons is 92.
• Radioactive decay: The spontaneous breakdown of such an unstable atomic nucleus causes radioactivity.
• The process of radioactive decay as a function of time is represented by

$ln (\frac{N}{N_0}) = - λ t$

N is the number of radioactive nuclei present in a sample, No is the number of radioactive nuclei present in the sample at t = 0, λ is disintegration constant.

• Disintegration constant: The disintegration constant is the unique constant related to the radioactive nature of the given nucleus.
• Half-Life Period: The time in which the number of nuclei reached to half of its initial value is called the half-life period.

It is the time period when N = N0 / 2

The half - Life period is given as

$ln (\frac{N_0 /2}{N_0}) = - λ t_{1/2}$

⇒ $t_{1/2} = \frac{ln2}{λ}$

Calculation:

Given half-life is 22 years

$\implies 22 = \frac{ln2}{λ}$

$\implies \lambda = \frac{ln2}{22}$ -- (1)

Now, we got the value of disintegration constant.

It is given that the substance is decreased to 10 %

So, N = 10 % of No

Or

$\frac{N}{N_0} = \frac{1}{10}$ --- (2)

We know that

$ln (\frac{N}{N_0}) = - λ t$

Using (1) and (2) in the above equation

$\implies ln (\frac{1}{10}) = - \frac{ln 2}{22} t$

$\implies - ln \ 10 = - \frac{ln 2}{22} t$

$\implies t = \frac{ln 10}{ln 2} \times 22$

⇒ t = 73.11 years.

So, the correct option is 73.11 years