Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2\, MeV$. A $\gamma $ -ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma $ -ray. If $E = B$, show that this cannot happen. Hence calculate how much bigger than $B$ must $E$ be for such a process to happen.
Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2\, MeV$. A $\gamma $ -ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma $ -ray. If $E = B$, show that this cannot happen. Hence calculate how much bigger than $B$ must $E$ be for such a process to happen.
Here, when deuteron nucleus is given energy $\mathrm{E}=\mathrm{B}$, proton and neutron (say particle $1$ and particle $2$) become free from mutual binding but since energy supplied is getting consumed completely, no energy remains in balance and so they can not be emitted out. This fact can be proved as follows.
Suppose after spending energy $B$ out of supplied energy $E$, proton and neutron (after getting freed from mutual binding) move with momenta, say $p_{1}$ and $p_{2}$ respectively and possess kinetic energy say $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively. If this would happen then,
$\mathrm{E}-\mathrm{B}=\mathrm{K}_{1}+\mathrm{K}_{2}$
... $(1)$
$\therefore \quad \mathrm{E}-\mathrm{B}=\frac{p_{1}^{2}}{2 m}+\frac{p_{2}^{2}}{2 m}$$....(2)$
(Where $m_{p} \approx m_{n} \approx m$ )
According to law of conservation of momentum, $p_{1}+p_{2}=p^{\prime} \quad\left(\right.$ As per the statement $\left.\overrightarrow{p^{\prime}}\left\|\overrightarrow{p_{1}}\right\| \overrightarrow{p_{2}}\right)$ Here momentum of photon,
Here momentum of photon,
$p^{\prime}=\frac{h}{\lambda}=\frac{h}{(c / f)}=\frac{h f}{c}=\frac{\mathrm{E}}{c}$$...(3)$
and so from above equation $p_{1}+p_{2}=\frac{\mathrm{E}}{c}$$...(4)$
If $E=B$ then from equation $(1)$,
$\frac{p_{1}^{2}}{2 m}+\frac{p_{2}^{2}}{2 m}=0$ $\therefore p_{1}^{2}+p_{2}^{2}=0$ $\Rightarrow p_{1}=0 \text { and } p_{2}=0$ $\Rightarrow\text { Proton and neutron can not be emitted, after being freed from mutual binding. }$
Now suppose, $\mathrm{E}=\mathrm{B}+\Delta \mathrm{E}$ (Where $\Delta \mathrm{E}<<\mathrm{E}$ )
$\therefore \mathrm{E}-\mathrm{B}=\Delta \mathrm{E}$$....(5)$
Similar Questions
A small quantity of solution containing $Na^{24}$ radio nuclide of activity $1$ microcurie is injected into the blood of a person. A sample of the blood of volume $1\, cm^3$ taken after $5$ hours shows an activity of $296$ disintegration per minute. What will be the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person ......... $liter$
(Take $1$ curie $= 3.7 × 10^{10}$ disintegration per second and ${e^{ - \lambda t}} = 0.7927;$ where $\lambda$= disintegration constant)
A small quantity of solution containing $Na^{24}$ radio nuclide of activity $1$ microcurie is injected into the blood of a person. A sample of the blood of volume $1\, cm^3$ taken after $5$ hours shows an activity of $296$ disintegration per minute. What will be the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person ......... $liter$
(Take $1$ curie $= 3.7 × 10^{10}$ disintegration per second and ${e^{ - \lambda t}} = 0.7927;$ where $\lambda$= disintegration constant)
Assertion : ${}^{90}Sr$ from the radioactive fall out from a nuclear bomb ends up in the bones of human beings through the milk consumed by them. It causes impairment of the production of red blood cells.
Reason : The energetic $\beta - $ particles emitted in the decay of ${}^{90}Sr$ damage the bone marrow
Assertion : ${}^{90}Sr$ from the radioactive fall out from a nuclear bomb ends up in the bones of human beings through the milk consumed by them. It causes impairment of the production of red blood cells.
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- [AIIMS 2004]
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- [KVPY 2012]
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