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A radioactive substance has a half life of 60 minutes. After 3 hours, the fraction of atom that have decayed would be

12.5%

87.5%

8.5%

25.1%

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Explanation

(b) $N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$ Hence fraction of atoms decayed
= $1 - \frac{N}{N_{0}} = 1 - {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} = 1 - {(\frac{1}{2})}^{\frac{3 \times 60}{60}} = \frac{7}{8}$
In percentage it is $\frac{7}{8} \times 100 = 87.5 %$

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An element used for radioactive carbon dating for more than 5600 years is

C-14

U-234

U-238

Po-94

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Explanation

(a) C-14 is carbon dating substance.

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N atoms of a radioactive element emit n alpha particles per second. The half life of the element is

$\frac{n}{N}$ sec

$\frac{N}{n}$ sec

$\frac{0.693 N}{n}$ sec

$\frac{0.693 n}{N}$ sec

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Explanation

$∴ λ = \frac{- n}{N}$ Half life = $\frac{0.693}{λ} = \frac{0.693 N}{n} \sec$

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The radioactive nucleus ${}_{7}^{13}N$ decays to ${}_{6}^{13}C$ through the emission of

Neutron

Proton

Electron

Positron

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Explanation

(d) ${}_{7}^{13}N \to {}_{6}^{13}C + {}_{+ 1}^{0}e$

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The radioactive decay of uranium into thorium is expressed by the equation ${}_{92}^{238}U \to {}_{90}^{234}{Th} + X$ where 'X' is

An electron

A proton

A deuteron

An alpha particle

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Explanation

(d) ${}_{92}^{238}U \to {}_{90}^{234}{Th} + {}_{2}^{4}{He}$

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In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following would be emitted in the decay process

Proton

Neutron

Electron

Photon

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Explanation

(d) In photon decay the mass no. and atomic mass remain the same.

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The half life (T) and the disintegration constant $λ$ of a radioactive substance are related as

$λT = 1$

$λT = 0.693$

$\frac{T}{λ} = 0.693$

$\frac{λ}{T} = 0.693$

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Explanation

(b) By using $N = N_{0} e^{- λt} \Rightarrow \frac{N_{0}}{2} = N_{0} e^{- {λT}_{1 / 2}} \Rightarrow 2 = e^{{λT}_{1 / 2}}$
By taking $\log_{e}$ both the side
$\log_{e} 2 = {λT}_{1 / 2} \Rightarrow {λT}_{1 / 2} = 0.693$

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The half life period of a radioactive substance is 5 min. The amount of substance decayed in 20 min will be

93.75%

75%

25%

6.25%

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Explanation

(a) Number of half lives in 20 min = n = $\frac{20}{5} = 4$
Fraction of material remains after four half lives = $\frac{1}{16}$
Hence fraction that decays $1 - \frac{1}{16} = \frac{15}{16} = 93.75 %$

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In the uranium radioactive series, the initial nucleus is ${}_{92}^{238}U$ and the final nucleus is ${}_{82}^{206}{Pb}$ . When the uranium nucleus decays to lead, the number of $α$ -particles emitted will be

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Explanation

(d) $n_{α} = \frac{A - A'}{4} = \frac{238 - 206}{4} = 8$

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Which of the following is a correct statement

Beta rays are same as cathode rays

Gamma rays are high energy neutrons

Alpha particles are singly ionized helium atoms

Protons and neutrons have exactly the same mass

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Explanation

(a) Both the $β$ -rays and the cathode rays are made up of electrons. $γ$ -rays are EM waves, $α$ -particles are doubly ionized helium atoms and protons and neutrons have approximately the same mass.

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