Chapter 28, Problem 66P

(a)

To determine

The speed of the selected neutrons escaping the reactor.

Answer to Problem 66P

The speed of the neutrons is $2.21\times {10}^{-34}\text{ m}$.

Explanation of Solution

The time interval between the two shutters is $13.0\text{ms}$ and the distance between the first and second shutter is $16.4\text{ m}$.

Write the expression for speed.

$v=\frac{x}{t}$                                                                          (I)

Here, $v$ is the speed of the neutron, $x$ is the distance travelled and $t$ is the time of travel.

Substituting  $13.0\text{ms}$ for $t$ and $16.4\text{ m}$ for $x$ in (I) to find $v$

$\begin{array}{c}v=\frac{16.4\text{ m}}{13.0\text{ms}}\\ =\frac{16.4\text{ m}}{13.0\times {10}^{-3}\text{s}}\\ =1.26\times {10}^3{\text{ms}}^{-1}\\ =1.26{\text{ kms}}^{-1}\end{array}$

Thus, the speed of the neutrons is $1.26{\text{ kms}}^{-1}$.

(b)

To determine

The de Broglie wavelength of the neutrons.

Answer to Problem 66P

The de Broglie wavelength of the neutrons is $\text{314 pm}$.

Explanation of Solution

The velocity of the neutrons is $1.26\times {10}^{-3}{\text{ms}}^{-1}$.

Write the expression for de Broglie wavelength

$\lambda =\frac{h}{p}$                                           (II)

Here, $\lambda$ is the de Broglie wavelength, $h$ is the Planck’s constant  and $p$ is the momentum of the bullet.

Substitute $p=mv$ in (I)

$\lambda =\frac{h}{mv}$                                           (III)

Here, $m$ is the mass of the neutrons and $v$ is its velocity.

Substitute $1.675\times {10}^{-27}\text{ kg}$ for $m$, $1.26\times {10}^{-3}{\text{ms}}^{-1}$ for $v$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{1.675\times {10}^{-27}\text{ kg}\times 1.26\times {10}^{-3}{\text{ms}}^{-1}}\\ =314\times {10}^{-12}\text{ m}\\ =\text{314 pm}\end{array}$

Thus, the de Broglie wavelength of the neutrons is $\text{314 pm}$.

(c)

To determine

The range of de Broglie wavelength.

Answer to Problem 66P

The range of de Broglie wavelength is $\text{303 pm to 324 pm}$.

Explanation of Solution

The slowest neutron takes longer time to get through the shutters. It is given by the sum of the time interval between the two shutters and the time the two shutters are open i.e.

$t_s=t+t_1$                                                                  (IV)

Here, $t_s$ is the time of travel of the slowest neutron and $t_1$ is the opening time of the shutter.

However, the fastest neutron takes lesser time and it is given by the difference of the time interval between the two shutters and the time the two shutters are open i.e.

$t_f=t-t_1$                                                                (V)

Here, $t_f$ is the time of travel of the fastest neutron.

Substitute $0.45\text{}\text{ ms}$ for $t_1$  and $13.0\text{ms}$ for $t$ in (IV)

$\begin{array}{c}{t}_s=0.45\text{}\text{ ms}+13.0\text{ms}\\ \text{=13}\text{.45}\text{ms}\end{array}$

Substitute $\text{13}\text{.45}\text{ms}$ for $t$ and $16.4\text{ m}$ for $x$ in (I) to find $v$

$\begin{array}{c}v=\frac{16.4\text{ m}}{13.45\text{ms}}\\ =\frac{16.4\text{ m}}{13.45\times {10}^{-3}\text{s}}\\ =1220{\text{ms}}^{-1}\end{array}$

Thus, the speed of the slowest neutrons is $1220{\text{ms}}^{-1}$.

Substitute $1.675\times {10}^{-27}\text{ kg}$ for $m$, $1.26\times {10}^{-3}{\text{ms}}^{-1}$ for $v$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{1.675\times {10}^{-27}\text{ kg}\times 1220{\text{ms}}^{-1}}\\ =324\times {10}^{-12}\text{ m}\\ =\text{324 pm}\end{array}$

Thus, the de Broglie wavelength of the fastest neutrons is $\text{324 pm}$.

Substitute $0.45\text{}\text{ ms}$ for $t_1$  and $13.0\text{ms}$ for $t$ in (V)

$\begin{array}{c}{t}_f=0.45\text{}\text{ ms}-13.0\text{ms}\\ \text{=12}\text{.55}\text{ms}\end{array}$

Substitute $\text{12}\text{.55}\text{ms}$ for $t$ and $16.4\text{ m}$ for $x$ in (I) to find $v$

$\begin{array}{c}v=\frac{16.4\text{ m}}{\text{12}\text{.55}\text{ms}}\\ =\frac{16.4\text{ m}}{12.55\times {10}^{-3}\text{s}}\\ =1307{\text{ms}}^{-1}\end{array}$

Thus, the speed of the fastest neutrons is $1307{\text{ms}}^{-1}$.

Substitute $1.675\times {10}^{-27}\text{ kg}$ for $m$, $1307{\text{ms}}^{-1}$ for $v$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{1.675\times {10}^{-27}\text{ kg}\times 1307{\text{ms}}^{-1}}\\ =303\times {10}^{-12}\text{ m}\\ =\text{303 pm}\end{array}$

Thus, the de Broglie wavelength of the slowest neutrons is $\text{303 pm}$.

Thus, the range of de Broglie wavelength is $\text{303 pm to 324 pm}$.