CBSE Class 12 · 2016 · All India · Modern PhysicsShort Answer

WhatsApp Email

(i) Derive the mathematical expression for the law of radioactive decay for a sample of radioactive nuclei.
(ii) How is the mean life of a given radioactive nucleus related to the decay constant?

Answer: $N = N_{0} e^{- λ t}$ ; mean life $τ = 1/ λ$ .

Solution

(i) The rate of decay is proportional to the number of undecayed nuclei:

\frac{d N}{d t} = - λ N \Rightarrow \frac{d N}{N} = - λ d t .

Integrating from

N_{0}

(at

t = 0

) to

N

(at time

t

\int_{N_{0}}^{N} \frac{d N}{N} = - λ \int_{0}^{t} d t \Rightarrow ln \frac{N}{N _{0}} = - λ t .

N = N_{0} e^{- λ t}

(ii) The mean (average) life is the reciprocal of the decay constant:

τ = \frac{1}{λ} .

CBSE Class 12 Board · 2016 · All India · Subjective

Practise more: all CBSE board PYQs on Modern Physics →

More board questions on Modern Physics

The potential energy between two nucleons inside a nucleus is minimum at a distance of about
CBSE 2024View solution →
The variation of the stopping potential $V_{0}$ with the frequency $ν$ of the incident radiation for four metals A, B, C and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will …
CBSE 2024View solution →
The energy of an electron in the ground state of a hydrogen atom is $- 13.6$ eV. The kinetic and potential energy of the electron in the first excited state will be
CBSE 2024View solution →
Assertion (A): An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of $18 0^{\circ}$ .
Reason (R): The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus.
CBSE 2024View solution →
Calculate the energy released / absorbed (in MeV) in the nuclear reaction $_{1}^{1} H +_{1}^{3} H \to_{1}^{2} H +_{1}^{2} H$ . Given $m (_{1}^{1} H) = 1.007825 u$ , $m (_{1}^{2} H) = 2.014102 u$ , …
CBSE 2024View solution →
A proton of energy 1.6 MeV approaches a gold nucleus $(Z = 79)$ . Find the distance of its closest approach.
CBSE 2024View solution →