Photons of wavelength $\lambda$ are incident on the cathode of a photocell. Electrons are emitted from the cathode surface. The de-Broglie wavelength of the emitted electrons is (work function is negligible).
($c =$ velocity of light,$h =$ Planck's constant,$m =$ mass of electron)
($c =$ velocity of light,$h =$ Planck's constant,$m =$ mass of electron)
- A$\sqrt{\frac{mc}{2h\lambda}}$
- B$\sqrt{\frac{h\lambda}{2mc}}$
- C$\sqrt{\frac{2h\lambda}{mc}}$
- D$\sqrt{\frac{mh}{\lambda c}}$
Explore More
Similar Questions
The kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:
An electron is accelerated from rest through a potential difference such that its kinetic energy is $1.5 \ eV$. The de-Broglie wavelength associated with this electron is:
Medium
View SolutionThe energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be
Difficult
View SolutionIf the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is
An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma$. The rate of change of de Broglie wavelength associated with the electron varies inversely as the $n^{\text{th}}$ power of time. The numerical value of $n$ is . . . . . . .
DifficultJEE MAIN 2025
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