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Physics NEET Questions — Free MCQ Practice

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Half-life of a radioactive substance is 12.5 h and its mass is 256 g. After what time, the amount of remaining substance is 1 g? [2001]

A.

75 h

B.

100 h

C.

125 h

D.

150 h

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Explanation

The mass of radioactive substance remained is,

$M = M_{0} {(\frac{1}{2})}^{n}$

Here, final mass, M = 1 g, initial mass, $M_{0}$ = 256 g, half life period, $T_{1 / 2}$ = 12.5 h

$s o, 1 = 256 {(\frac{1}{2})}^{n} o r \frac{1}{256} = {(\frac{1}{2})}^{n} o r {(\frac{1}{2})}^{8} = {(\frac{1}{2})}^{n} c o m p a r i n g t h e p o w e r s o n b o t h t h e s i d e s, w e g e t n = 8 = \frac{t}{T_{1 / 2}} ∴ t = 8 T_{1 / 2} = 8 \times 12.5 = 100 h$

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A radioactive substance disintegrates 1/64 of initial value in 60 s. The half-life of this substance is

A.

5 s

B.

10 s

C.

30 s

D.

20 s

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Explanation

$\frac{1}{64} = {(\frac{1}{2})}^{n} n = 6$

6 half-lives are equal to 60 second. $∴ 1 h a l f - l i f e = 10 s$

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The nucleus ${}_{6}C_{12}$ absorbs an energetic neutron and emits a beta particle $(β^{-})$ . The resulting nucleus is

A.

${}_{7}N_{14}$

B.

${}_{7}N_{13}$

C.

${}_{5}B_{13}$

D.

${}_{6}C_{13}$

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Explanation

A nuclear reaction represents the transformation of one stable nucleus into another nucleus by bombarding the former with suitable high energy particles.

${}_{6}C_{12} + {}_{0}n_{1} \to {}_{6}C_{13} \to {}_{7}N_{13} + {}_{- 1}β_{0} + Q (e n e r g y)$

Resulting nucleus is of nitrogen having mass no. 13 and atomic no. 7

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If in a nuclear fusion process, the masses of the fusion nuclei be $m_{1} a n d m_{2}$ and the mass of the resultant nucleus be $m_{3}$ , then [2004]

A.

$m_{3} = m_{1} + m_{2}$

B.

$m_{3} = |m_{1} - m_{2}|$

C.

$m_{3} < (m_{1} + m_{2})$

D.

$m_{3} > (m_{1} + m_{2})$

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Explanation

In a nuclear fusion, when two light nuclei of different masses are combined to form a stable nucleus, then some mass is lost and appears in the form of energy, called the mass defect. So, the mass of resultant nucleus is always less than the sum of masses of initial nuclei i.e.,

$m_{3} < (m_{1} + m_{2})$

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The nuclei of which one of the following pairs of nuclei are isotones? [2005]

A.

${}_{34}S e^{74}, {}_{31}G a^{71}$

B.

${}_{42}M o^{92}, {}_{40}Z r^{92}$

C.

${}_{38}S r^{84}, {}_{38}S r^{86}$

D.

${}_{20}C a^{40}, {}_{16}S_{32}$

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Explanation

The nuclei which have same number of neutrons but different atomic number and mass number are known as isotones. In choice (1) nuclei of ${}_{34}S e^{74}$ and ${}_{31}G a^{71}$ are isotones as

A - Z = 74 - 34 = 71 - 31 = 40

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The counting rate observed from a radio active source at t = 0 second was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The counting rate observed, as counts per second, at t = 6 seconds will be

A.

400

B.

300

C.

200

D.

150

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Explanation

$A = |\frac{- d N}{d t}| = c o u n t i n g r a t e = a c t i v i t y = N λ A_{0} = 1600 a t t = 0 A = 100 a t t = 8 s \frac{A}{A_{0}} = \frac{100}{1600} = \frac{1}{16} = {(\frac{1}{2})}^{44} = \frac{N λ}{N_{0} λ} = \frac{N}{N_{0}} ∴ \frac{N}{N_{0}} = {(\frac{1}{2})}^{4} = {(\frac{1}{2})}^{t / T} ∴ \frac{t}{T} = 4, t = 8 s; T = h a l f l i f e = 2 s; t' = 6 s e c o n d = 3 h a l f l i f e ∴ \frac{A'}{A_{0}} = \frac{N' λ}{N_{0} λ} = \frac{N'}{N_{0}} = {(\frac{1}{2})}^{3} = \frac{1}{8} \frac{A'}{1600} = \frac{1}{8}; A' = \frac{1600}{8} = 200$

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If $N_{0}$ is the original mass of the substance of half life period $T_{1 / 2}$ = 5 years, then the amount of substance left after 15 years is [AIEEE 2012]

A.

$N_{0}$ /8

B.

$N_{0}$ /16

C.

$N_{0}$ /2

D.

$N_{0}$ /4

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Explanation

$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} = N_{0} {(\frac{1}{2})}^{\frac{15}{5}} = \frac{N_{0}}{8}$

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The numbers of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time t = 2 s. Then number of nuclei at time t = 4 s will be

A.

800

B.

810

C.

790

D.

700

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Explanation

In 2 s only 90% nuclei are left behind. Thus, in the next 2 s, 90% of 900 or 810 nuclei will be left.

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A nucleus ${}_{Z}^{A}X$ has mass represented by m(A, Z). If $m_{p}$ and $m_{n}$ denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then [2007]

A.

BE = [m(A, Z) - Z $m_{p}$ - (A - Z) $m_{n}$ ] $c^{2}$

B.

BE = [Z $m_{p}$ + (A - Z) $m_{n}$ - m(A,Z)] $c^{2}$

C.

BE = [Z $m_{p}$ + A $m_{n}$ - m(A, Z)] $c^{2}$

D.

BE = m(A, Z) - Z $m_{p}$ - (A, Z) $m_{n}$

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Explanation

In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to disappearance of a fraction of the total mass. If the quantity of mass disappearing is $∆$ m, then the binding energy is BE = $∆$ m $c^{2}$

From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the consituent neutrons and protons. We can then write.

$∆ m = Z m_{p} + N m_{n} - m (A, Z)$

where m(A, Z) is the mass of the atom of mass number A and atomic number Z. Hence, the binding energy of the nucleus is

BE = [Z $m_{p}$ + N $m_{n}$ - m(A, Z) $c^{2}$

BE = [Z $m_{p}$ + (A-Z) $m_{n}$ - m(A, Z) $c^{2}$

where, N = A -Z = number of neutrons

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The equation ${}_{Z}X_{A} \to {}_{Z + 1}Y_{A} + {}_{- 1}e_{0} + \bar{ν}$ is (UP CPMT 2002)

A.

$β$ -emission

B.

$α$ -emission

C.

$e^{-}$ capture

D.

fission

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Explanation

The equation ${}_{Z}X_{A} \to {}_{Z + 1}Y_{A} + {}_{- 1}e_{0} + \bar{ν}$ represents $β$ -emission (since $β$ -ray electron is emitted)

Physics MCQs for NEET — Practice Questions with Answers

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