Question
Electrons are accelerated through a potential difference of $\mathrm{V}_{0} \mathrm{kV}$. The de Broglie wavelength (in $\mathrm{m}$ ) of the electrons is(a) $\frac{\mathrm{h}}{\sqrt{1000 \mathrm{meV}_{0}}}$(b) $\frac{\mathrm{h}}{\sqrt{2000 \mathrm{meV}_{0}}}$(c) $\frac{\mathrm{h}}{\sqrt{\mathrm{meV}_{0}}}$(d) $\frac{\mathrm{h}}{2 \mathrm{meV}_{0}}$
Step 1
Step 1: The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$, where $h$ is the Planck's constant, $m$ is the mass of the electron and $K$ is the kinetic energy of the electron. Show more…
Show all steps
Your feedback will help us improve your experience
Ajay Singhal and 76 other Physics 103 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
An electron of mass $m$ when accelerated through a potential difference has de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be (b) $\lambda \sqrt{\frac{m}{M}}$ (a) $\lambda \frac{m}{M}$ (c) $\lambda \frac{M}{m}$ (d) $\lambda \sqrt{\frac{M}{m}}$
Dual Nature of Radiation and Matter
Round 2
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential difference. the ratio of the de Broglie wavelength associated with an electron to that associated with proton is (a) 1 (b) $m_{p} / m_{e}$ (c) $m_{e} / m_{p}$ (d) $\sqrt{m_{p} / m_{e}}$
Electrons are accelerated by a $1000-\mathrm{V}$ potential drop. $(a)$ Calculate the de Broglie wavelength. (b) Calculate the wavelength of the $X$ -rays that could be produced when these electrons strike a solid.
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD