Physics Nuclei

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New answer posted•a year ago

Physics Nuclei Class 12th

A radioactive $β -$ emission. A detector records $n β -$ particles in $2 s$ and by next $2 s$ (accumulatively) it records $1.1 n β -$ particle. Number of $β -$ particles recorded by detector after a long time, is

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$I_{S_{4}} = 4 I_{0}$

$I_{S_{3}} = I_{0}$

$∴ \frac{I_{m a x}}{I_{m i n}} = \frac{{(\sqrt[]{I_{S_{4}}} + \sqrt[]{I_{S_{3}}})}^{2}}{{(\sqrt[]{I_{S_{4}}} - \sqrt[]{I_{S_{3}}})}^{2}} = {(\frac{2 + 1}{2 - 1})}^{2} = \frac{9}{1}$

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New answer posted•a year ago

Physics Nuclei Class 12th

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

27 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Physics Nuclei Class 12th

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.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Energy of products = 939 + 940 = 1879
ΔE + E_reactions – E_productor = 1877 – 1879 = -2MeV.
Hence it must capture a γ - ray photon of energy 2MeV.

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New answer posted•a year ago

Physics Nuclei Class 12th

The binding energy (in MeV) of a nitrogen nucleus $(_{7}^{14} N)$ is :

(Given mass of nucleus $m (_{7}^{14} N) = 14.00307 u$ , mass of proton $m_{p} = 1.00783 u$ , mass of neutron $m_{n} = 1.00867 u$ and $1 u = 931.5 M e V / c^{2}$ )

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$_{7}^{14} N$ contains 7 protons and 7 neutrons.

Mass defect, $(_{}_{}) (\begin{matrix} \end{matrix})$ $Δ m = (7 m_{H} + 7 m_{n}) - m (\begin{matrix} 14 \\ 7 \end{matrix}) = (7 \times 1.00783 u) + (7 \times 1.00867 u) - 14.00307 u$

$= 0.11243 u$

Binding energy $= 0.11243 \times 931.5 M e V = 104.7 M e V$

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New answer posted•a year ago

Physics Nuclei Class 12th

A radioactive sample decays with a constant of (1/3)logₑ2 s⁻¹. If initially there are 200 nuclei present, find the number of nuclei decayed during the first 9 seconds.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

t? , half life = (log?2)/λ = 3 s
Initially 200 nuclei were there
After 3 half-lives, 9 seconds,
No. of nuclei = (1/2³) (200) = 25
Therefore, the number of nuclei that decayed = 200 – 25 = 175

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New answer posted•a year ago

Physics Nuclei Class 12th