The energy of a photon is $E = hv$ and the momentum of photon $p = \frac{h}{\lambda }$, then the velocity of photon will be
$E/p$
$Ep$
${\left( {\frac{E}{p}} \right)^2}$
$3 \times {10^8}m/s$
By suitable example show that particle or wave nature depend on type of experiment.
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 radio transmitter radiates $1 kW$ power at a wavelength $198.6 $ metres. How many photons does it emit per second
The time taken by a photoelectron to come Out after the photon strikes is approximately
The minimum intensity of light to be detected by human eye is ${10^{ - 10}}W/{m^2}$. The number of photons of wavelength $5.6 \times {10^{ - 7}}m$ entering the eye, with pupil area ${10^{ - 6}}{m^2}$, per second for vision will be nearly