A photon of $1.7 \times {10^{ - 13}}$ Joules is absorbed by a material under special circumstances. The correct statement is
A photon of $1.7 \times {10^{ - 13}}$ Joules is absorbed by a material under special circumstances. The correct statement is
- A
Electrons of the atom of absorbed material will go the higher energy states
- B
Electron and positron pair will be created
- C
Only positron will be produced
- D
Photoelectric effect will occur and electron will be produced
Similar Questions
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is $10.2 \ eV$. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of $15 \ eV$. What will be observed by the detector
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is $10.2 \ eV$. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of $15 \ eV$. What will be observed by the detector
- [IIT 2005]
The photoelectric effect can be understood on the basis of
The photoelectric effect can be understood on the basis of
If the energy of the photon is increased by a factor of $4$, then its momentum
If the energy of the photon is increased by a factor of $4$, then its momentum
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\, \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3}$ $watt$ will be : $\left(\mathrm{h}=6.6 \times 10^{-34}\, \mathrm{Js}\right)$
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\, \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3}$ $watt$ will be : $\left(\mathrm{h}=6.6 \times 10^{-34}\, \mathrm{Js}\right)$
- [NEET 2021]
A radiation of energy $'E'$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $( C =$ Velocity of light $)$
A radiation of energy $'E'$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $( C =$ Velocity of light $)$
- [AIPMT 2015]