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Which of the following pairs is an isobar

${}_{1}H_{1}$ and ${}_{1}H_{2}$

${}_{1}H_{2}$ and ${}_{1}H_{3}$

${}_{6}C_{12}$ and ${}_{6}C_{13}$

${}_{15}P_{30}$ and ${}_{14}{Si}_{30}$

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Explanation

(d) Isobars have equal mass numbers.

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Equivalent energy of mass equal to 1 a.m.u. is

931 KeV

931 eV

931 MeV

9.31 MeV

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Explanation

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The binding energies per nucleon for a deuteron and an particle are $x_{1}$ and $x_{2}$ respectively. What will be the energy Q released in the reaction ${}_{1}H_{2} + {}_{1}H_{2} \to {}_{2}{He}_{4} + Q$

$4 (x_{1} + x_{2})$

$4 (x_{2} - x_{1})$

$2 (x_{1} + x_{2})$

$2 (x_{2} - x_{1})$

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Explanation

(b) Q = $4 (x_{2} - x_{1})$

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The rest energy of an electron is

510 KeV

931 KeV

510 MeV

931 MeV

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Explanation

(a) Rest energy of an electron = $m_{e} c^{2}$
Here $m_{e} = 9.1 \times 10^{- 31} kg$ and c = velocity of light
$∴$ Rest energy = $9.1 \times 10^{- 31} \times {(3 \times 10^{8})}^{2}$ joule
$= \frac{9.1 \times 10^{- 31} \times (3 \times 10^{8})^{2}}{1.6 \times 10^{- 19}} eV = 510 keV$

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In ${}_{88}{Ra}_{226}$ nucleus, there are

138 protons and 88 neutrons

138 neutrons and 88 protons

226 protons and 88 electrons

226 neutrons and 138 electrons

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Explanation

(b) ${}_{z}X_{A} = {}_{88}{Ra}_{226}$
Number of protons = Z = 88
Number of neutrons = A-Z = 226-88 = 138.

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Outside a nucleus

Neutron is stable

Proton and neutron both are stable

Neutron is unstable

Neither neutron nor proton is stable

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Explanation

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Order of magnitude of density of uranium nucleus is ( $m_{P} = 1.67 \times 10^{- 27} kg$ )

$10^{20} kg / m^{3}$

$10^{17} kg / m^{3}$

$10^{14} kg / m^{3}$

$10^{11} kg / m^{3}$

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Explanation

(b) The order of magnitude of mass and volume of uranium nucleus will be
m â‰ƒ A( $1.67 \times 10^{- 27}$ kg) (A is atomic number)
$V = \frac{4}{3} {πr}^{3} ≃ \frac{4}{3} π {[(1.25 \times 10^{- 15} m) A^{1 / 3}]}^{3} ≃ (8.2 \times 10^{- 45} m^{3}) A$

Hence, $ρ = \frac{m}{V} = \frac{A (1.67 \times 10^{- 27} kg)}{(8.2 \times 10^{- 45} m^{3}) A}$

$≃ 2.0 \times 10^{17} kg / m^{3}$
.

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Radius of ${}_{2}^{4}{He}$ nucleus is 3 Fermi. The radius of ${}_{82}^{206}{Pb}$ nucleus will be

5 Fermi

6 Fermi

11.16 Fermi

8 Fermi

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Explanation

(c) We have $r \propto A^{1 / 3} \Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{206}{4})}^{1 / 3}$

$∴ r_{2} = 3 {(\frac{206}{4})}^{1 / 3} = 11.16 Fermi$

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Nucleus of an atom whose atomic mass is 24 consists of

11 electrons, 11 protons and 13 neutrons

11 electrons, 13 protons and 11 neutrons

11 protons and 13 neutrons

11 protons and 13 electrons

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Explanation

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The mass of a neutron is the same as that of

A proton

A meson

An epsilon

An electron

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Explanation

(a) Mass of proton and neutron is lame.

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