ATOMS AND NUCLEI

6 )

Nucleon is common name for

electron and neutron

proton and neutron

neutron and positron

neutron and neutrino

View Solution

Answer :

Nucleon means neutron and proton.

7 )

The ratio of atomic volume of nuclear volume is of the order of

10^(â€“15)

10^(â€“10)

10^(15)

10^(10)

View Solution

Answer :

[(volume of atom) / (volume of nucleus)] = [{(4 / 3)πr_a³} / {(4 / 3)πr_n³}]

= (r_a / r_n)³

= [(10^–10) / (10^–15)]³

= (10⁵)³

= 10¹⁵.

8 )

The size of the nucleus is of the order of

10^(â€“10) m

10^(â€“14) m

10^(14) m

10^(4) m

View Solution

Answer :

size of nucleus is the order 10^–14 m to 10^–15 m.

9 )

The size of the atom is of the order of

10^(â€“14) m

10^(â€“10) m

10^(â€“4) m

10^(â€“6) m

View Solution

Answer :

size of atom is the order 10^–10 m.

10 )

The half time of a radioactive substance is 20 min, difference between Points of time when it is 33% disintegrated and 67% disintegrated is approximately

10 min

20 min

24 min

28 min

View Solution

Answer :

amount of fraction left = (N / N₀) and

(N / N₀) = (1 / 2)^{[t / {T(1/2)}]}

where T_(1/2) = half life = 20 min

∴ (N / N₀) = (1 / 2)^{(t / 20)}

(i) when 33 % is disintegrated i.e. 67 % is remained hence

[(67) / (100)] = (1 / 2)^{[{t(1)} / (20)]} (1)

(ii) when 67 % disintegrated i.e. 33 % is removed hence

[(33) / (100)] = (1 / 2)^{[{t(2)} / (20)]} (2)

from (1) & (2)

(67 / 33) = (1 / 2)^{[{t(1) – t(2)} / (20)]}

∴ 2.03 = (1 / 2)^{[{t(1) – t(2)} / (20)]}

∴ log 2.03 = [(t₁ – t₂) / (20)] log (1 / 2)

∴ [(t₁ – t₂) / (20)] = [(log 2.03) / (log 0.5)]

∴ t₁ – t₂ = 20 × (– 1.021)

t₂ – t₁ = 20.4 min

t₂ – t₁ ≈ 20 min

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