Question

1. set the electron gun energy between 1 and 4 mev. find a slit spacing that gives 7 to 11 well - defined diffraction fringes. then, by using the laser and video camera, find the wavelength that gives this same diffraction pattern for this particular slit spacing. repeat this procedure for two different kinetic energies in that range of mev, and record the values in the following table. then calculate the constant that relates p with λ.

electron kinetic energy	slit spacing	diffraction pattern	wavelength	constant
3 mev	187 nm	9 fringes	20.7 nm	6.63·10⁻³⁴ j·s
4 mev

Explanation:

Step1: Recall the formula for diffraction

For a diffraction - grating type situation with electrons or light, the formula for the position of the maxima in a diffraction pattern is \(d\sin\theta = n\lambda\), where \(d\) is the slit - spacing, \(\theta\) is the angle of diffraction, \(n\) is the order of the maximum (\(n = 1,2,\cdots\)), and \(\lambda\) is the wavelength. In the case of electrons, the de - Broglie wavelength is given by \(\lambda=\frac{h}{p}\), where \(h = 6.63\times10^{-34}\ J\cdot s\) is Planck's constant and \(p\) is the momentum of the electron.

Step2: For the first row with \(E = 2\ meV\)

First, convert the electron kinetic energy from \(meV\) to \(J\). We know that \(1\ meV=1\times10^{- 3}\ eV\) and \(1\ eV = 1.6\times10^{-19}\ J\), so \(E = 2\times10^{-3}\ eV=2\times10^{-3}\times1.6\times10^{-19}\ J = 3.2\times10^{-22}\ J\). The kinetic energy of an electron is \(E=\frac{p^{2}}{2m}\), where \(m = 9.11\times10^{-31}\ kg\) is the mass of the electron. Then \(p=\sqrt{2mE}=\sqrt{2\times9.11\times10^{-31}\times3.2\times10^{-22}}\ kg\cdot m/s\approx7.61\times10^{-27}\ kg\cdot m/s\). The de - Broglie wavelength \(\lambda=\frac{h}{p}\), and we are given \(\lambda = 31.7\ nm=31.7\times10^{-9}\ m\). Using the diffraction formula \(d\sin\theta = n\lambda\), assuming a small - angle approximation (\(\sin\theta\approx\tan\theta\) for small \(\theta\)) and for \(n = 9\) (the number of fringes), we can find \(d\). Rearranging \(d=\frac{n\lambda}{\sin\theta}\), in the small - angle limit, if we assume a simple setup, we can use the relation for the first - order approximation. But if we just use the formula \(d=\frac{n\lambda}{1}\) (since we are not given the angle details and for a basic understanding), \(d = 9\times31.7\ nm=285.3\ nm\)

Answer:

285.3