$A$ photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)
- A$\sqrt{E / 2m}$
- B$c \sqrt{2mE}$
- C$c \sqrt{\frac{2m}{E}}$
- D$\frac{1}{c} \sqrt{E / 2m}$
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An electron (mass $m$) with initial velocity $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $\overrightarrow{E} = -E_{0} \hat{k}$. If $\lambda_{0}$ is the initial de-Broglie wavelength of the electron,its de-Broglie wavelength at time $t$ is given by:
DifficultJEE MAIN 2020
View Solution$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.
Medium
View SolutionThe equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:
For which of the following particles will it be most difficult to experimentally verify the de Broglie relationship?
The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \ m$ when its kinetic energy is $1.0 \ keV$. Its wavelength will be $1.0 \times 10^{-10} \ m$ when its kinetic energy is: (in $keV$)
MediumWBJEE 2017
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