| ⇦ |
| ⇨ |
The radioactivity of a certain material drops to 1/16 of the initial value in 2 hours. The half-life of this radio nuclide is
Options
(a) 10 min
(b) 20 min
(c) 30 min
(d) 40 min
Correct Answer:
30 min
Explanation:
In 4 × T₁/₂, N = (N₀ / 2) × (1 / 2) × (1 / 2) × (1 / 2)
= N₀ / 16 4 × T₁/₂ = 2 hrs.
⇒ T₁/₂ = 30 min
Related Questions: - In a circuit, L,C and R are connected in series with an alternating voltage source
- One-fourth length of a spring of force constant K is cut away. The force constant
- The time of reverberation of a room A is one second. What will be the time (in seconds)
- The lowest frequency of light that will cause the emission of photoelectrons
- In a mass spectrometer used for measuring the masses of ions, the ions
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a circuit, L,C and R are connected in series with an alternating voltage source
- One-fourth length of a spring of force constant K is cut away. The force constant
- The time of reverberation of a room A is one second. What will be the time (in seconds)
- The lowest frequency of light that will cause the emission of photoelectrons
- In a mass spectrometer used for measuring the masses of ions, the ions
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply