New Notes Blog: A neutron in a reactor has kinetic energy of approximately 0.020 eV. Calculate the wavelength of this neutron.

Sunday, 11 August 2013

A neutron in a reactor has kinetic energy of approximately 0.020 eV. Calculate the wavelength of this neutron.

The de Broglie relation is

$\displaystyle{e}{q}.{\left({1}\right)}\to$ $\displaystyle{P}=\frac{{h}}{\lambda}$

Now get the momentum $\displaystyle{P}$ in terms of the kinetic energy $\displaystyle{K}_{{E}}$ .

$\displaystyle{K}_{{E}}=\frac{{1}}{{2}}{m}{{v}}^{{2}}$

$\displaystyle{K}_{{E}}=\frac{{{P}}^{{2}}}{{{2}{m}}}$

$\displaystyle{e}{q}.{\left({2}\right)}\to$ $\displaystyle{P}=\sqrt{{{2}{m}{K}_{{E}}}}$

Therefore the wavelength can be found by equating $\displaystyle{e}{q}.{\left({2}\right)}$ and $\displaystyle{e}{q}.{\left({1}\right)}$ and solving for $\displaystyle\lambda$ .

$\displaystyle\frac{{h}}{\lambda}=\sqrt{{{2}{m}{K}_{{E}}}}$

$\displaystyle\frac{\lambda}{{h}}=\frac{{1}}{\sqrt{{{2}{m}{K}_{{E}}}}}$

$\displaystyle\lambda=\frac{{h}}{\sqrt{{{2}{m}{K}_{{E}}}}}$

Plug in numerical values. You will find the units are much easier if you convert the neutron mass into energy with units of $\displaystyle{e}{V}$ by $\displaystyle\ldots$

The de Broglie relation is

$\displaystyle{e}{q}.{\left({1}\right)}\to$ $\displaystyle{P}=\frac{{h}}{\lambda}$

Now get the momentum $\displaystyle{P}$ in terms of the kinetic energy $\displaystyle{K}_{{E}}$ .

$\displaystyle{K}_{{E}}=\frac{{1}}{{2}}{m}{{v}}^{{2}}$

$\displaystyle{K}_{{E}}=\frac{{{P}}^{{2}}}{{{2}{m}}}$

$\displaystyle{e}{q}.{\left({2}\right)}\to$ $\displaystyle{P}=\sqrt{{{2}{m}{K}_{{E}}}}$

Therefore the wavelength can be found by equating $\displaystyle{e}{q}.{\left({2}\right)}$ and $\displaystyle{e}{q}.{\left({1}\right)}$ and solving for $\displaystyle\lambda$ .

$\displaystyle\frac{{h}}{\lambda}=\sqrt{{{2}{m}{K}_{{E}}}}$

$\displaystyle\frac{\lambda}{{h}}=\frac{{1}}{\sqrt{{{2}{m}{K}_{{E}}}}}$

$\displaystyle\lambda=\frac{{h}}{\sqrt{{{2}{m}{K}_{{E}}}}}$

Plug in numerical values. You will find the units are much easier if you convert the neutron mass into energy with units of $\displaystyle{e}{V}$ by $\displaystyle{E}={m}{{c}}^{{2}}$ . Hence,

$\displaystyle\lambda=\frac{{h}}{\sqrt{{{2}{m}{{c}}^{{2}}{K}_{{E}}}}}$

$\displaystyle\lambda=\frac{{{1240}{e}{V}\cdot{n}{m}}}{\sqrt{{{2}{\left({940}\cdot{{10}}^{{6}}{e}{V}\right)}{\left({0.020}{e}{V}\right)}}}}={0.20}{n}{m}$

New Notes Blog

Sunday, 11 August 2013

A neutron in a reactor has kinetic energy of approximately 0.020 eV. Calculate the wavelength of this neutron.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Sunday, 11 August 2013

A neutron in a reactor has kinetic energy of approximately 0.020 eV. Calculate the wavelength of this neutron.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?