Home
Class 12
CHEMISTRY
The half-life period of U^(234) is 2.5 x...

The half-life period of `U^(234)` is `2.5 xx 10^(5)` years. In how much is the quantity of the isotope reduce to 25% of the original amount?

Text Solution

Verified by Experts

Initial amount of this isotope `N_(0) = 100`
Final amount of the isotop `N = 25`
We know that `N = ((1)/(2))^(n) N_(0)`
So `25 = ((1)/(2))^(n) xx 100`
or `(25)/(100) = ((1)/(2))^(n)`
or `(25)/(100) = ((1)/(2))^(n)`
or `((1)/(2))^(2) = ((1)/(2))^(n)`
or `n = 2`
`= T = n xx t_(1//2) = 2 xx 2.5 xx 10^(5) = 5 xx 10^(5)` year
Alternatively
We can use `(t_(1//2))/(t_(x%)) = (0.3)/("log"(a)/(a - x))`
`= (t_(1//2))/(t_(25%)) = (0.3)/("log"(100)/(25))`
`(2.5 xx 10^(5))/(t_(25%)) = (0.3)/("log"4)`
`t_(25%)` or time required to reduced amount to 25%
`= (2.5 xx 10^(5))/(0.3) xx "log" 4`
`= 2.5 xx (10^(5))/(0.3) xx 0.602`
`= 5.0 xx 10^(5)` years
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY|Exercise Solved Example|18 Videos

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY|Exercise Ex6.1 Objective|15 Videos

  • NCERT BASED EXERCISE

    CENGAGE CHEMISTRY|Exercise Nuclear Chemistry (NCERT Exercise)|29 Videos

  • ORGANIC COMPOUNDS WITH FUNCTIONAL GROUP

    CENGAGE CHEMISTRY|Exercise Archives Analytical And Descriptive|24 Videos

Similar Questions

Explore conceptually related problems

Radioactive disintegration is a first order reaction and it's rate depends only upon the nature of nucleus and does not depend upon external factors like temoperature an pressure. The rate of radioactive disintigration (Activity) is respresented as (-d(N))/(dt)=lambdaN Where lambda =decay consatant, N= number of nuclei at time t, N""_(0) =initial no. of nucleei. The above equation after integration can be represented as lamda=2.303/tlog.(N""_(0))/(N) Half-life period of U""^(237) is 2.5xx10""^(5) years. In how much time will the amount of U""^(237) remaining be only 25% of the original amount?

Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Half-life period of U^(2.5xx10^(5) years. In how much thime will the amount of U^(237) remaining be only 25% of the original amount ?

The half -life period of ._(84)Po^(210) is 140 days. In how many days 1 g of this isotope is reduced to 0.25 g ?

The half-life period of a radioactive element is 1.4 xx 10^(10) years. Calculate the time in which the activity of the element is reduced to 90% of its original value.

The half life period of a first order reaction, A to Product is 10 minutes. In how much time is the concentration of A reduced to 10% of its original concentration?

The mass of 1 curie of U-234 (t_(1//2)=2*35 xx 10^(5) "years") is

The half life period of radium is 1600 years . In how many years 1 g radium will be reduced to 0.125 g ?

For a first order reactions, the half -life is 10 mins. How much time in minutes will it take to reduce the concentration of reactant to 25% of its original concentration ?