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If a radioactive material reduces to 12.5% of its value in 24 days, then its half-life is:

16 days

8 days

4days

2 days

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Explanation

$(2) N = N_{0} {(\frac{1}{2})}^{n} \Rightarrow \frac{12.5}{100} N_{0} = N_{0} {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{n} \Rightarrow n = 3 T_{H} = \frac{24}{n} = \frac{24}{3} = 8 days$

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The half-life of a radioactive material depends on

The initial amount of material

Rate of disintegration

Decay constant

Temperature and pressure

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Explanation

$(3) T_{H} = \frac{0.693}{λ}$

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Binding energy per nucleon is:

Maximum for middle-order elements

Maximum for lighter order elements

Maximum for heavier order elements

None of these

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Explanation

1. It is maximum for middle-order elements.

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$In the equation {}_{13}^{27}{AI} + {}_{2}^{4}{He} \to {}_{15}^{30}P + X where X is -$

${}_{1}^{1}P$

${}_{1}^{2}H$

${}_{2}^{4}{He}$

${}_{0}^{1}n$

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Explanation

$(4) {}_{z}X^{A} = \{A = (27 + 4 - 30) = 1 Z = (13 + 2) - 15 = 0\} \Rightarrow {}_{0}X^{1}$

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The half-life of a radioactivity isotope X is 100years. It decays to an element Y which is stable. The two elements are found to be ratio 1: 31 in a sample of a given rock. Assuming that at the time of formation of the rock there was only isotope X, then estimate the age of the rock.

300 years

400 years

500 years

600 years

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Explanation

$(3) N = N_{0} {(\frac{1}{2})}^{n} \frac{1}{32} = {(\frac{1}{2})}^{n} \Rightarrow n = 5 t = {nT}_{H} = 100 x 5 = 500 yrs$

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A chain reaction is continuous due to

Large mass defect

Large energy

Production of more neutrons in fission

None of these

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Explanation

3. Production of more neutrons in fission

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Density (D) of nucleus related to mass no. (A) as

$D \propto A$

$D \propto A^{2}$

$D \propto A^{0}$

$D \propto A^{- 1}$

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Explanation

$(3) Density = \frac{mass}{volume} = \frac{A M_{p}}{\frac{4}{3} {πR}_{0}^{3} A} [M_{p} : M a s s o f a n u c l e o n] S o d e n s i t y i s i n d e p e n d e n t o f A$

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Heavier nuclei are unstable. The correct statement is :

binding energy per nucleon is small

get stability by nuclear fission

get stability by nuclear fusion

both (1) and (2)

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Explanation

Fon heavier nuclei, binding energy per nucleon is minimum.

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$binding energy per nucleon for deuteron ({}_{1}H_{2}) and helium ({}_{2}{He}_{4}) are 1.1 MeV and 7.0 MeV respectively . Energy released when two deuteron fuse to form a He nucleus is$

47.12 MeV

34.4 MeV

11.8 MeV

23.6 MeV

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Explanation

$(4) {}_{1}H_{2} + {}_{1}H_{2} \to {}_{2}{He}_{2} B . E . of reactant = 2 x 1.1 x 2 = 4.4 MeV B . E . of product = 7.0 x 4 = 28.0 MeV Energy released = 28.0 - 4.4 = 23.6 MeV$

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The initial activity of a certain radioactive isotope was measured as 16000 counts per minute. That the only activity measured was due to this and that its activity after 12 hours was 2000 counts/minute. Its half-life, in hours, is nearest to-

9.0

6.0

4.0

3.0

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Explanation

$(3) 2000 = 16000 {(\frac{1}{2})}^{n} n = 3 T_{H} = \frac{12}{3} = 4$

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