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Physics NEET Questions — Free MCQ Practice

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Mean life of a radioactive sample is 100 seconds. Then its half life (in minutes) is

A.

0.693

B.

1

C.

$10^{- 4}$

D.

1.155

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Explanation

(d) Mean life (T) = 1/ $λ$ = 100 second
Half-life = $\frac{0.693}{λ} = \frac{0.693 \times 100}{60} = 1.155 \min$

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${}_{86}^{222}A \to {}_{84}^{210}B$ . In this reaction how many $α$ and $β$ particles are emitted

A.

6 $α$ , 3 $β$

B.

3 $α$ , 4 $β$

C.

4 $α$ , 3 $β$

D.

3 $α$ , 6 $β$

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Explanation

(b) By using $n_{α} = \frac{A - A'}{4}$ and $n_{β} = 2 n_{α} - Z + Z'$

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The phenomenon of radioactivity is

A.

Exothermic change which increases or decreases with temperature

B.

Increases on applied pressure

C.

Nuclear process does not depend on external factors

D.

None of the above

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Explanation

(c) The phenomenon of radioactivity does not depend on external factos.

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If half life of radium is 77 days. Its decay constant in day will be

A.

$3 \times 10^{- 13}$ /day

B.

$9 \times 10^{- 3}$ /day

C.

$1 \times 10^{- 3}$ /day

D.

$6 \times 10^{- 3}$ /day

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Explanation

(b) $λ = \frac{0.693}{T_{1 / 2}} = \frac{0.693}{77} = 9 \times 10^{- 3} / day$

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In a sample of radioactive material, what fraction of the initial number of active nuclei will remain undisintegrated after half of a half-life of the sample

A.

$\frac{1}{4}$

B.

$\frac{1}{2 \sqrt{2}}$

C.

$\frac{1}{\sqrt{2}}$

D.

$2 \sqrt{2}$

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Explanation

(c) $∴ \frac{N}{N_{0}} = {(\frac{1}{2})}^{t / T_{1 / 2}} {(\frac{1}{2})}^{1 / 2} = \frac{1}{\sqrt{2}}$

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Consider two nuclei of the same radioactive nuclide. One of the nuclei was created in a supernova explosion 5 billion years ago. The other was created in a nuclear reactor 5 minutes ago. The probability of decay during the next time is

A.

Different for each nuclei

B.

Nuclei created in explosion decays first

C.

Nuclei created in the reactor decays first

D.

Independent of the time of creation

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Explanation

(d) The half life and decay constant, independent of time of creation of radioactive nuclei.

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An $α$ -particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180^o. The nearest distance upto which $α$ -particle reaches the nucleus will be of the order of

A.

1 Å

B.

$10^{- 10} cm$

C.

$10^{- 12} cm$

D.

$10^{- 15} cm$

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Explanation

(c) At closest distance of approach
Kinetic energy = Potential energy

$\Rightarrow 5 \times 10^{6} \times 1.6 \times 10^{- 19} = \frac{1}{4 {πε}_{0}} \times \frac{(ze) (2 e)}{r}$

For uranium z = 92, so r = $5.3 \times 10^{- 12} cm$

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In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy $E_{0}$ and the radius $r_{0}$ of the first orbit will be ( $a_{0}$ is the Bohr radius)

A.

$E_{0} = - 27.2 eV; r_{0} = a_{0} / 2$

B.

$E_{0} = - 27.2 eV; r_{0} = a_{0}$

C.

$E_{0} = - 13.6 eV; r_{0} = a_{0} / 2$

D.

$E_{0} = - 13.6 eV; r_{0} = a_{0}$

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Explanation

(a) Here radius of electron orbit r ∝ 1/m and energy E ∝ m, where m is the mass of the electron.
Hence energy of hypothetical atom
$E_{0} = 2 \times (- 13.6 eV) = - 27.2 eV$ and radius $r_{0} = \frac{a_{0}}{2}$

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A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for exciting electron from first to third Bohr orbit in ${Li}^{+ +}$ will be (Ionisation energy of hydrogen atom is 13.6eV)

A.

182.51 Å

B.

177.17 Å

C.

142.25 Å

D.

113.74 Å

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Explanation

(d) $E_{n} = - 13.6 \frac{Z^{2}}{n^{2}} eV$

Required energy for said transition

$∆ E = E_{3} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$

$\Rightarrow ∆ E = 13.6 \times 3^{2} [\frac{8}{9}] = 108.8 eV$

$\Rightarrow ∆ E = 108.8 \times 1.6 \times 10^{- 19} J$

Now $∆ E = \frac{hc}{λ} = 108.8 \times 1.6 \times 10^{- 19}$

$\Rightarrow λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{108.8 \times 1.6 \times 10^{- 19}} = 0.11374 \times 10^{- 7} m = 113.74 \overset{0}{A}$

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The ionisation potential of H-atom is 13.6 V. When it is excited from ground state by monochromatic radiations of $970.6 \overset{0}{A}$ , the number of emission lines will be (according to Bohr’s theory)

A.

10

B.

8

C.

6

D.

4

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Explanation

(c) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

$\Rightarrow \frac{1}{970.6 \times 10^{- 10}} = 1.097 \times 10^{7} [\frac{1}{1^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow n_{2} = 4$

$∴$ Number of emission lines $N = \frac{n (n - 1)}{2} = \frac{4 \times 3}{2} = 6$

Physics MCQs for NEET — Practice Questions with Answers

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