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Physics MCQs for NEET — Practice Questions with Answers

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The average binding energy of a nucleon inside an atomic nucleus is about [1989]

8 MeV

8 eV

8 J

8 erg

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Explanation

Average binding energy per nucleon of a nucleus is the average enegy we have to spend to remove a nucleon from the nucleus to infinite distance. It is given by total binding energy divided by the mass number of the range A = 30 to A = 120 corresponding to average binding energy per nucleon = 8 MeV. The peak value of the maximum is 8 MeV / N for ${}_{26}F e^{56}$ .

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A nucleus with Z = 92 emits the following in a sequence:

$α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . The Z of the resulting nucleus is [AIEEE 2003]

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Explanation

$Z_{R e s u l t i n g n u c l e u s} = 92 - 8 \times 2 + 4 \times 1 - 2 \times 1 = 78$

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$Two nucleons are at a separation of 1 fm . The net force between them is F_{1} if both are neutrons, F_{2} if both are protons and F_{3} if one is proton and other is a neutron . Hence$

$F_{1} > F_{2} > F_{3}$

$F_{2} > F_{1} > F_{3}$

$F_{1} = F_{3} > F_{2}$

$F_{1} = F_{2} > F_{3}$

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Explanation

Nuclear force independent of charge but if the electrostatic force is considerable then force between neutrons and neutron-proton pair is greater because electrostatic repulsion is there among the protons

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$A nucleus {}_{Z}^{A}X emits one α - particle and one β - particle . The resulting nucleus is$

${}_{Z - 1}^{A - 2}X$

${}_{Z - 1}^{A - 4}X$

${}_{Z}^{A - 4}X$

${}_{Z - 2}^{A - 4}X$

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Explanation

$(2) {}_{Z}X_{4} \overset{α}{\to} {}_{Z - 2}X'^{(A - 4)} \overset{β}{\to} {}_{Z - 1}X^{(A - 4)}$

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The ratio of the radius of the nucleus to the radius of an atom is of the order of :

$10^{- 5}$

$10^{- 7}$

$10^{- 2}$

$10 °$

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Explanation

The radius of the nucleus is the order of fermi and atom is the order of angstrom.

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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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