Photon and electron are given same energy $({10^{ - 20}}J)$. Wavelength associated with photon and electron are ${\lambda _{Ph}}$ and ${\lambda _{el}}$ then correct statement will be
Photon and electron are given same energy $({10^{ - 20}}J)$. Wavelength associated with photon and electron are ${\lambda _{Ph}}$ and ${\lambda _{el}}$ then correct statement will be
- A
${\lambda _{Ph}} > {\lambda _{el}}$
- B
${\lambda _{Ph}} < {\lambda _{el}}$
- C
${\lambda _{Ph}} = {\lambda _{el}}$
- D
$\frac{{{\lambda _{el}}}}{{{\lambda _{Ph}}}} = C$
Similar Questions
Monochromatic light of wavelength $667 \,\,nm$ is produced by a helium neon laser. The power emitted is $9 \,\,mW.$ The number of photons arriving per sec. on the average at a target irradiated by this beam is
Monochromatic light of wavelength $667 \,\,nm$ is produced by a helium neon laser. The power emitted is $9 \,\,mW.$ The number of photons arriving per sec. on the average at a target irradiated by this beam is
- [AIPMT 2009]
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy $10.2\; BeV$ into two $\gamma$ -rays of equal energy. What is the wavelength associated with each $\gamma$ -ray? $\left(1\; BeV =10^{9}\; eV \right)$
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy $10.2\; BeV$ into two $\gamma$ -rays of equal energy. What is the wavelength associated with each $\gamma$ -ray? $\left(1\; BeV =10^{9}\; eV \right)$
If we express the energy of a photon in $KeV$ and the wavelength in angstroms, then energy of a photon can be calculated from the relation
The electrons are emitted in the photoelectric effect from a metal surface
The electrons are emitted in the photoelectric effect from a metal surface