Question

A beam of electrons of $\mathbf{\text{25\hspace{0.17em}eV}}$kinetic energy is directed at a single slip of 2.0 μm width, then detected at a screen 4m beyond the slit. How far from a point directly in the line of the beam is the first location where no electrons are ever detected?

Short answer

The line of the beam is detected at$\text{y}=\text{0.25mm}$

Step-by-step solution

 Step 1: Given information

The potential difference for the electron beam is$\text{25\hspace{0.17em}eV}$

The slit's width$\text{=2\hspace{0.17em}μm}$

The space between the screen and the slit$=4\text{\hspace{0.17em}m}$

Concept Introduction

The wavelength of any moving charged particle is given by the,

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{\text{h}}}{\sqrt{\mathbf{\text{2mqV}}}}$…………………..(1)

Where m, q, and V are the mass, charge, and applied potential difference.

Calculation of the wavelength

Substitute the given information in equation (1) to obtain the wavelength such that,

$\begin{array}{c}\lambda =\frac{6.63\times {10}^{-34}J\cdot s}{\sqrt{2\times (9.1\times {10}^{-31}kg)\times (1.6\times {10}^{-19}C)\times \left(25V\right)}}\\ =2.46\times {10}^{-10}m\end{array}\frac{}{}$

 Another formula substituting.

$\begin{array}{l}m\lambda =\omega \sin \theta \\ \frac{\text{1}}{\text{2}}\lambda =\text{ω}\times \frac{\text{y}}{\text{D}}\\ \frac{\text{1}}{\text{2}}\times \text{2.46}\times {\text{10}}^{\text{-10}}m=\text{2.0}\times {\text{10}}^{-6}m\times \frac{\text{y}}{4m}\\ y=0.25mm\end{array}$

Therefore, the result is y=0.25 mm.