Physics Nuclei Overview

Determine the number of nuclei that will be decayed upto one average life of radioactive material? (N₀ : Initial number of nuclei)

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N = N₀e^–^λ^t

->t = λN = N₀/e

So number of nuclei decayed = N₀ – N

= N₀ (1- 1/e)

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Density of nucleus is of the order of

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Density of nuclei is order of 10¹⁷kg m^–3.

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Power factor of series LCR circuit at resonance is

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Density of nucleus is of the order of

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Density of nuclei is order of 10¹⁷ kg m^-³.

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Two radioactive substances X and Y originally have N₁ and N₂ nuclei respectively. Half life of X is half of the life of Y. After three half lives of Y, number of nuclei of both are equal.

The ratio $\frac{N_{1}}{N_{2}}$ will be equal to:

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Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_{1}}{2^{6}} = \frac{N_{2}}{2^{3}} \Rightarrow \frac{N_{1}}{N_{2}} = 8$

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The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6/25% of its initial value. The value of x is___________.

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According to Nuclear activity, we can write

$N_{0} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{2} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{4} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{8} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{16} = (0.0625) N_{0}$

Time required = 4 * $t_{\frac{1}{2}} = 20 y r s$

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5 months ago

In the following unclear reaction,

D $\overset{α}{\to} D_{1} \overset{β^{-}}{\to} D_{2} \overset{α}{\to} D_{3} \overset{γ}{\to} D_{4}$

Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D₄ respectively will be________

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A ray of light is incident at an angle of incidence $60 °$ on the glass slab of refractive index $\sqrt[]{3} .$ After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $4 \sqrt[]{3}$ cm. The thickness of the glass slab is _______cm.

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$d = 4 \sqrt[]{3}$ cm (Lateral shift)

By Snell's law

$μ_{a i r} s i n 60 ° = μ_{g} s i n θ$

θ = 30

$\Rightarrow 1 * \frac{\sqrt[]{3}}{2} = \sqrt[]{3} s i n θ$

$s i n θ = \frac{1}{2}$

t = 12 cm

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5 months ago

Mass of Bromine, Selenium and Krypton are given as:-

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$Q_{(i)} = (M_{B r} - M_{K r}) c^{2} > 0$

$\begin{array}{l} Q_{(i i)} = (M_{B r} - M_{S e} - 2 M_{e}) c^{2} > 0 \\ Q_{(i i)} = (M_{B r} - M_{s e}) c^{2} > 0 \end{array}$

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The rest mass of the deuteron, proton and neutron is equivalent to energy of 1877MeV, 939MeV, 940MeV respectively. A deuteron may disintegrate to a proton and neutron if it.

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