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2 months ago

Physics Nuclei Class 12th

Following statements related to radioactivity are given below:

(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.

(C) Slope of the graph of log_e (no. of undecayed nuclei) Vs, time represents the reciprocal of mean life time $(τ) .$

(D) Product of decay constant $(λ)$ and half-life time $(T_{1 / 2})$

...more

Following statements related to radioactivity are given below:

(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.

(C) Slope of the graph of log_e (no. of undecayed nuclei) Vs, time represents the reciprocal of mean life time $(τ) .$

(D) Product of decay constant $(λ)$ and half-life time $(T_{1 / 2})$ is not constant.

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alok kumar singh

Contributor-Level 10

We know, N = $N_{0} e^{- λ t}$

Where, N Number of un-decayed Nuclei

No Initial No. Of Nuclei

N $e^{- λ t}$

$N = N_{0} {\bar{e}}^{λ t}$

Taking log both side log N = log N0 + loge $e^{- λ t}$

log N = log No $λ t$

Slope = $λ = \frac{1}{t_{a v}}$

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New answer posted

2 months ago

Physics Nuclei Class 12th

$\sqrt[]{d_{1}} a n d \sqrt[]{d_{2}}$ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an a particle is approaching a gold nucleus. For d₁ = x d₂, the value of x will be________.

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alok kumar singh

Contributor-Level 10

$\sqrt[]{d}$ $c o t \frac{θ}{2}$

cot² 30° = x cot² 45°

x = 3

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2 months ago

Physics Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The activity of a radioactive material is 2.56 * 10^-³ Ci. If the half life of the material is 5 days, after how many days the activity will become 2 * 10^-⁵ Ci?

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alok kumar singh

Contributor-Level 10

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

&nbs

...more

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 * t_1/2

= 7 * 5 = 35 days

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Physics Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ${}_{1}^{2}X +_{1}^{2} X = \frac{4}{2} Y$

The binding energies per nucleon for are 1.1 Me V and 7.6

Me V respectively. The energy released in this process is________ Me V.

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alok kumar singh

Contributor-Level 10

Energy released = Binding energy of product – Binding energy of reacts :

$Δ E = 7.6 * 4 - (1.1 * 4)$

= 26 MeV

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