$A$ 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$\frac{1}{c} \sqrt{E / 2m}$
Explore More
Similar Questions
$A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles,$\lambda_1 / \lambda_2$ is
MediumIIT 1999
View SolutionThe de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)
An electron is moving with initial velocity $\vec{v} = V_0 \hat{j}$ in a magnetic field $\vec{B} = B_0 \hat{i}$. Its de-Broglie wavelength will
An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by
Difficult
View SolutionThe de-Broglie wavelength is proportional to
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo