Nuclei
- In a radioactive material the activity at time t1 is R1 and at a later time t2, it is R2. If the decay constant of the material is λ, then
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Let at time t1 & t2, number of particles be N1 & N2. So,
R1 = dN1 = -λN1; R2 = dN2 = -λ N2 dt dt R1 = λN1 = N1 = e -λ(t2 - t1) R2 λN2 N1e -λ(t2 - t1)
R1 = R2eλ(t2 - t1) = R2e-λ(t2 - t1)Correct Option: D
Let at time t1 & t2, number of particles be N1 & N2. So,
R1 = dN1 = -λN1; R2 = dN2 = -λ N2 dt dt R1 = λN1 = N1 = e -λ(t2 - t1) R2 λN2 N1e -λ(t2 - t1)
R1 = R2eλ(t2 - t1) = R2e-λ(t2 - t1)
- Alpha-particles are
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We know that alpha particles are the nucleus of ionized helium atoms which contain two protons and two neutrons. These are emitted by the nuclei of certain radioactive substances. Streams of alpha particles, called α-rays, produce intense ionisation in gases through which they pass and are easily absorbed by matter.
Correct Option: D
We know that alpha particles are the nucleus of ionized helium atoms which contain two protons and two neutrons. These are emitted by the nuclei of certain radioactive substances. Streams of alpha particles, called α-rays, produce intense ionisation in gases through which they pass and are easily absorbed by matter.
- A sample has 4 × 1016 radioactive nuclei of half life 10 days. The number of atoms decaying in 30 days is
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N = 4 × 1016 1 30/10 = 1 × 1016 2 2
Atoms decayed= 4 × 1016 - 1 × 1016 2
= 3.5 × 1016Correct Option: D
N = 4 × 1016 1 30/10 = 1 × 1016 2 2
Atoms decayed= 4 × 1016 - 1 × 1016 2
= 3.5 × 1016
- A deuteron strikes 8O16 nucleus with subseq-uent emission of an alpha particle. Identify the nucleus so produced
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8O16 + 1H2 → 2He4 + 7N14
Correct Option: D
8O16 + 1H2 → 2He4 + 7N14
- The decay constant (λ) and the half-life (T) of a radioactive isotope are related as
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t = 1 log a λ a - x
When t = T,x = a 2 T = 1 log a = 1 loge2 λ a - a λ 2 ⇒ λ = 1 loge2 T Correct Option: A
t = 1 log a λ a - x
When t = T,x = a 2 T = 1 log a = 1 loge2 λ a - a λ 2 ⇒ λ = 1 loge2 T