Question

Neutron diffraction is used in determining the structures of molecules.

a. Calculate the de Broglie wavelength of a neutron moving at$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\%}$of the speed of light.

b. Calculate the velocity of a neutron with a wavelength of $\mathbf{75}\mathit{p}\mathit{m}\left(1pm={10}^{-12}m\right)$

Short answer

Answer

a) de Broglie wavelength for neutron is $1.32\times {10}^{-13}m$.

b) Velocity for neutron is $5.3\times {10}^{3}m/s$.

Step-by-step solution

Concept used

de-Broglie relation:

The wave associated with matter can be linked to particle characteristics. This fact was observed by Louis de Broglie. The relationship amongst mass, velocity, and wavelength is established as follows:

$\lambda =hmv$

Where,

$\lambda$denotes wavelength.

$m$denotes mass.

$v$ denotes velocity.

$h$ denotes Planck's constant.

Calculate de Broglie wavelength for neutron

a)

The velocity of the neutron is $1\%$that of the velocity of light.

Value of $c$is$3\times {10}^{8}m/s$

So, the velocity of the neutron is calculated as follows:

Velocity of neutron$\left(1\%\right)\left(3\times {108}^{8}m/s\right)-3\times {10}^{6}m/s$

The formula to calculate wavelength as per de Broglie relation is as follows:

$\lambda =hmv$ ……. (1)

Value of h is $6.626\times {10}^{-34}Js$.

Value of $m$is $1,67\times {10}^{-27}kg$.

Value of $v$is $3\times {10}^{6}m/s$.

Substitute the value in equation (1),

$\lambda =hmv$

$$\begin{gathered} =6.626\times {10}^{-34}Js\left(1.67\times {10}^{-27}kg\right)\left(3\times {10}^{6}m/s\right) \\ =1.32\times {10}^{-13}m \end{gathered}$$

Thus, de Broglie wavelength for neutron is $1.32\times {10}^{-13}m$.

Calculate the velocity for neutron

b)

De Broglie wavelength for neutron is 75pm.

The formula to calculate velocity from wavelength as per de Broglie relation is as follows:

$v=hm\lambda$ ……. (2)

Value of h is $6.626\times \times {10}^{-34}Js$.

Value of m is $1.67\times {10}^{-27}kg$.

Value of $\lambda$is $75pm$.

Substitute the value in equation (2),

$v=hm\lambda$

$$\begin{gathered} =6.626\times {10}^{-34}Js\left(1.67\times {10}^{-27}kg\right)\left(75pm\right)\left(1\times {10}^{-12}m1pm\right) \\ =5.29\times \times {10}^{3}m/s \\ \approx 5.3\times {10}^{3}m/s \end{gathered}$$

Thus, velocity for neutron is $5.3\times {10}^{3}m/s$.