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NEET NEET Questions — Free MCQ Practice

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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 $N = \frac{n (n - 1)}{2} = \frac{4 \times 3}{2} = 6$

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A neutron with velocity V strikes a stationary deuterium atom. Its kinetic energy changes by a factor of

A.

$\frac{15}{16}$

B.

$\frac{1}{2}$

C.

$\frac{2}{1}$

D.

None of these

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The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is 1.4 $kilowatt / m^{2}$ .The average earth- sun distance is $1.5 \times 10^{11}$ metres. The mass lost by the sun per day is
(1 day = 86400 seconds)

A.

$4.4 \times 10^{9} kg$

B.

$7.6 \times 10^{14} kg$

C.

$3.8 \times 10^{12} kg$

D.

$3.8 \times 10^{14} kg$

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Explanation

(d) Energy radiated = 1.4 $kW / m^{2}$

$= 1.4 kJ / \sec m^{2} = \frac{1.4 kJ}{\frac{1}{86400} day m^{2}} = \frac{1.4 \times 86400}{day m^{2}}$

Total energy radiated/day

$= \frac{4 π \times {(1.5 \times 10^{11})}^{2} \times 1.4 \times 86400}{1} \frac{kJ}{day} = E$

$∴ E = {mc}^{2} \Rightarrow m = \frac{E}{c^{2}}$

$= \frac{4 π \times {(1.5 \times 10^{11})}^{2} \times 1.4 \times 86400}{{(3 \times 10^{8})}^{2}} = 3.8 \times 10^{14} kg$

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The binding energy per nucleon of $O^{16}$ is 7.97 MeV and that of $O^{17}$ is 7.75 MeV. The energy (in MeV) required to remove a neutron from $O^{17}$ is

A.

3.52

B.

3.64

C.

4.23

D.

7.86

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Explanation

(c) The equation is $O^{17} \to {}_{0}^{1}n + O^{16}$
$∴$ Energy required = B.E. of $O^{17}$ – B.E. of $O^{16}$
= 17 $\times$ 7.75 – 16 $\times$ 7.97 = 4.23 MeV

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The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5 c. The change in its energy will be

A.

0.026 MeV

B.

0.051 MeV

C.

0.079 MeV

D.

0.105 MeV

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Explanation

(c) $∆ = {mc}^{2} - m_{0} c^{2} = \frac{m_{0} c^{2}}{\sqrt{1 - (v^{2} / c^{2})}} - m_{0} c^{2}$

$= m_{0} c^{2} (\frac{1}{\sqrt{1 - (v^{2} / c^{2})}} - 1) = 0.511 (\frac{1}{\sqrt{0.75}} - 1)$

= 0.079 MeV

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