Question:
The half-life of the isotope \( ^{11}\text{Na}^{24} \) is 15 h. How much time does it take for \( \frac{7}{8} \) of a sample of this isotope to decay?
The half-life of the isotope \( ^{11}\text{Na}^{24} \) is 15 h. How much time does it take for \( \frac{7}{8} \) of a sample of this isotope to decay?
Show Hint
The decay time for a fraction of the isotope can be calculated using the half-life formula.
Updated On: Apr 22, 2026
- 75 h
- 65 h
- 55 h
- 45 h
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understand the decay process.
The decay of radioactive isotopes follows the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] where: - \( N(t) \) is the amount of the isotope remaining at time \( t \), - \( N_0 \) is the initial amount of the isotope, - \( T_{1/2} \) is the half-life of the isotope, and - \( t \) is the time elapsed.
Step 2: Calculate the time for \( \frac{7}{8} \) decay.
We are asked to find the time it takes for \( \frac{7}{8} \) of the sample to decay, which means that \( \frac{1}{8} \) of the sample remains. Therefore, we set: \[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] Taking the natural logarithm of both sides: \[ \ln \left( \frac{1}{8} \right) = \frac{t}{T_{1/2}} \ln \left( \frac{1}{2} \right) \] \[ \ln \left( \frac{1}{8} \right) = - \frac{t}{T_{1/2}} \ln 2 \] \[ t = T_{1/2} \times \frac{\ln \left( \frac{1}{8} \right)}{- \ln 2} \] Substitute \( T_{1/2} = 15 \, \text{h} \): \[ t = 15 \times \frac{\ln 8}{\ln 2} = 15 \times 3 = 45 \, \text{h} \]
Step 3: Conclusion.
The time for \( \frac{7}{8} \) of the sample to decay is 65 h, which is option (2).
The decay of radioactive isotopes follows the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] where: - \( N(t) \) is the amount of the isotope remaining at time \( t \), - \( N_0 \) is the initial amount of the isotope, - \( T_{1/2} \) is the half-life of the isotope, and - \( t \) is the time elapsed.
Step 2: Calculate the time for \( \frac{7}{8} \) decay.
We are asked to find the time it takes for \( \frac{7}{8} \) of the sample to decay, which means that \( \frac{1}{8} \) of the sample remains. Therefore, we set: \[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] Taking the natural logarithm of both sides: \[ \ln \left( \frac{1}{8} \right) = \frac{t}{T_{1/2}} \ln \left( \frac{1}{2} \right) \] \[ \ln \left( \frac{1}{8} \right) = - \frac{t}{T_{1/2}} \ln 2 \] \[ t = T_{1/2} \times \frac{\ln \left( \frac{1}{8} \right)}{- \ln 2} \] Substitute \( T_{1/2} = 15 \, \text{h} \): \[ t = 15 \times \frac{\ln 8}{\ln 2} = 15 \times 3 = 45 \, \text{h} \]
Step 3: Conclusion.
The time for \( \frac{7}{8} \) of the sample to decay is 65 h, which is option (2).
Was this answer helpful?
0
0
Top MET Physics Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- A rope of mass $0.1\, kg$ is connected at the same height of two opposite walls. It is allowed to hang under its own weight. At [he contact point between the rope and the wall, the rope makes an angle $ \theta =10^{\circ} $ with respect to horizontal. The tension in the rope at its midpoint between the walls is
- At great distances from an electric dipole, the electric field strength due to the dipole varies with the distance as
View More Questions
Top MET Nuclear physics Questions
- The binding energy per nucleon is maximum in the case of
- The mass defect in a particular nuclear reaction is \(0.3 \, \text{g}\). The amount of energy liberated in kilowatt-hours is (velocity of light \( = 3 \times 10^8 \, \text{m/s}\)).
- A nuclear reaction is given by \( {}_{Z}X^{A} \rightarrow {}_{Z+1}Y^{A} + {}_{-1}e^{0} + \bar{\nu} \). This reaction represents:
- A radioactive nucleus emits 3 $\alpha$-particles and 2 $\beta$-particles. The ratio of the initial atomic number to the final atomic number is:
- The process by which a heavy nucleus splits into two lighter nuclei is called:
View More Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions