Recent content by sportfreak801

Thanks for the tip! That may be a possibility; however, we were provided what I had presumed to be two different samples. The first being the reference sample made up of known layers: a Silicon layer of 115\times10^{15} atoms/cm^{2} then a Bismuth layer of 5.65\times10^{15} atoms/cm^{2} and...

I took the Q's to be approximately equal to each other since the same dose was used in both the reference experiment and the example experiment.

I'll be more careful next time... It seems that the Nt for Beryllium is quite large compared to the Nt for Bismuth. There is a factor of 10^{3} difference, are such densities plausible?

I tried calculating the mass of the target particle of the first peak in my energy spectra M_{2} and then the number of particles per cm^2 Nt. However, the numbers that I found do not seem to be correct: K = \frac{E_{1}}{E_{0}}=\left ( \frac{M_{1}cos\theta + \sqrt{M_{2}^{2}-M_{1}^{2}\left (...

To calculate Z I would use the equation: \left ( \frac{d\sigma}{d\Omega} \right )=\left [ \frac{Z_{1}Z_{2}e^{2}}{4E} \right ]^{2}\frac{4\left \{ \left [ 1-\left ( \left ( M_{1}/M_{2} \right )sin^{2}\theta \right ) \right ]^{1/2} + cos\theta \right \}^{2}}{sin^{4}\theta \left [ 1-\left (...

Thank you for your response! I do, however, have some questions that I would like clarified. Does this mean that multiple peaks in the spectra correspond to the same element? For example, if in my spectra I have three distinct peaks that occur at different energy levels and have different...

In class we learned about the fundamentals behind RBS using a linear accelerator that accelerates He+ ions to 2MeV. We were then provided an example where those ions collided with a sample with layers of Silicon-Bismuth-Silicon, each of known thickness; and, we were then shown the corresponding...

\oint D\cdot nda = Q_e D\oint da = Q_e D(2\pi rl) = Q_e D = \frac{\lambda}{2r\pi} E_1 = \frac{D}{\epsilon_1} = \frac{\lambda}{2r\pi \epsilon_1} The potential from a to b over the first dielectric \Delta\varphi_1 = E_1 (b - a) \Delta\varphi_1 = \frac{\lambda...

Homework Statement A coaxial cable of circular cross section has a compound dielectric. The inner conductor has an outside radius a , which is surrounded by a dielectric sheath of dielectric constant K_1 and of outer radius b . Next comes another dielectric sheath of dielectric constant...

Now, I am supposed to derive the same answer using \varphi(r) = \frac{1}{4\pi\epsilon_0}\int\frac{dq}{|r - r'|} I know that q = \rho V and dq = \rho dV And that V = \frac{4}{3}\pi r^3 However, I know that dV = 4\pi r^2 dr is not true because if it were then I would get \varphi(r)...

My mistake, I forgot to substitute r = R... So it should read \frac{\rho}{\epsilon_0}\frac{R}{3} = -\frac{\alpha}{R^2} -\frac{\rho R^3}{3\epsilon_0} = \alpha -\frac{\rho R^2}{6\epsilon_0} + D = -\frac{\alpha}{R} -\frac{\rho R^2}{6\epsilon_0} + D = \frac{\rho R^2}{3\epsilon_0}...

-\mathbf{\nabla}((-\frac{\rho}{\epsilon_0}\frac{r^2}{6}) + D)= -\mathbf{\nabla}(-\frac{\alpha}{r}) -\frac{d}{dr}((-\frac{\rho}{\epsilon_0}\frac{r^2}{6}) + D)= -\frac{d}{dr}(-\frac{\alpha}{r}) \frac{\rho}{\epsilon_0}\frac{r}{3} = -\frac{\alpha}{r^2} -\frac{\rho r^3}{3\epsilon_0} = \alpha...

C = 0 So if the gradient of the potential must be continuous everywhere then -\mathbf{\nabla}((-\frac{\rho}{\epsilon_0}\frac{R^2}{6}) + D)= -\mathbf{\nabla}(-\frac{\alpha}{R})

So then the Electric Field is continuous everywhere. Also, the electric potential is finite everywhere. \textbf{E}=-\mathbf{\nabla}\varphi So I take the -\mathbf{\nabla}\varphi in spherical coordinates knowing that the Electric field is independent of \theta and \phi .

The Electric field \textbf{E}=-\mathbf{\nabla}\varphi is discontinuous when the electric field passes through the charged surface of the sphere.