Physics Nuclei

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a month ago

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R
Raj Pandey

Contributor-Level 9

11 22 N a X + e + + v 11 22 N a 10 22 N e + e + + v

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V
Vishal Baghel

Contributor-Level 10

Always possible as it is associated only with β? decay


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R
Raj Pandey

Contributor-Level 9

From Radioactive Decay Law,
dN/dt = λ? N + λ? N = λ_eff N
⇒ λ_eff = λ? + λ? ⇒ ln (2)/T = ln (2)/T? + ln (2)/T? ⇒ T = (T? ) / (T? + T? )
(where T, T? , and T? are half-lives)

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A
alok kumar singh

Contributor-Level 10

Total mass of reactant should be greater than that of product.
This condition is only fulfilled in case- 3

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alok kumar singh

Contributor-Level 10

The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction remaining is N'/N? = 3/4.

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alok kumar singh

Contributor-Level 10

P = E/t
= 2/235 * (6.023*10²? *200*1.6*10? ¹? )/ (30*24*60*60) = 60W

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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A
alok kumar singh

Contributor-Level 10

A = A 0 e - λ t

500 = 700 e - λ t λ t = l n ? 7 5 l n ? 2 t 1 / 2 * 30 = l n ? 7 5 ? t 1 / 2 = l n ? 2 * 30 l n ? 7 5

t 1 / 2 = 61.8 m i n 62 m i n .

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