Chapter 28, Problem 1P

To determine

Distance of interference.

Answer to Problem 1P

The answer is $\text{3}{\text{.6×10}}^{\text{-7}}\text{m}$

Explanation of Solution

Given info: Kinetic Energy, $\text{KE}\text{=}\frac{1}{40}\text{eV=}\text{0}\text{.025 eV}$

It is known that $\text{1}\text{eV}\text{=}\text{1}\text{.60}{\text{×10}}^{\text{-19}}\text{J/eV}$ and

Proton rest mass is $\text{1}\text{.67}{\text{×10}}^{\text{-27}}\text{kg}$

Distance (altitude) is $\text{1}\text{.0}\text{m}$

The neutron in a parallel beam are directed through two slits $\text{0}\text{.50}\text{mm}$ apart.

Formula used:

The wavelength is $\text{λ=}\frac{\text{h}}{\text{p}}\text{=}\frac{\text{h}}{{\left[{\text{2m}}_{\text{0}}\left(\text{KE}\right)\right]}^{\frac{\text{1}}{\text{2}}}}$

  $KE$ is kinetic energy.

  $m_0$ is rest mass

  $\text{λ}$ is wavelength

  $p$ is photon’s momentum

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

Calculation:

First calculate the wavelength of neutron by using the formula $\text{λ=}\frac{\text{h}}{{\left[{\text{2m}}_{\text{0}}\left(\text{KE}\right)\right]}^{\frac{\text{1}}{\text{2}}}}$

  $\text{=}\frac{\left(\text{6}{\text{.63×10}}^{\text{-34}}\text{J×s}\right)}{{\left[\text{2}\left(\text{1}{\text{.67×10}}^{\text{-27}}\text{kg}\right)\left(\text{0}\text{.025}\text{eV}\right)\left(\text{1}{\text{.60×10}}^{\text{-19}}\text{J/eV}\right)\right]}^{\frac{\text{1}}{\text{2}}}}\text{=1}{\text{.81×10}}^{\text{-10}}\text{m}$

Understand that the peaks of interference pattern are written as $\text{d}\text{sin}\text{θ}\text{=}\text{nλ,}\text{n =}\text{1,}\text{2,}\mathrm{......}$ and the positions on the screen are $\text{y = L}\text{tan}\text{θ}\text{.}$

For small angles $\text{sin}\text{θ=}\text{tan}\text{θ}$ So, $\text{y = }\frac{\text{n}\text{L}\text{λ}}{\text{d}}\text{.}$

Thus, the separation is $\Delta \text{y = }\frac{\text{L}\text{λ}}{\text{d}}\mathrm{...}\left(1\right)$

Substitute the all the values in equation $\left(1\right)$

  $\text{y = }\frac{\left(\text{1}\text{.0}\text{m}\right)\left(\text{1}{\text{.81×10}}^{\text{-10}}\text{m}\right)}{\left(\text{0}\text{.50}\text{m}\right)}\text{=}\text{3}{\text{.6×10}}^{\text{-7}}\text{m}$

Conclusion:

The required value is $\text{3}{\text{.6×10}}^{\text{-7}}\text{m}$ .