The photoelectric effect can be understood on the basis of

The momentum of a photon of energy $1 \text{ MeV}$ in $\text{kg-m/s}$ will be

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? $(1 \; BeV = 10^9 \; eV)$

$10^{20}$ photons of wavelength $660 \ nm$ are emitted per second from a lamp. The wattage of the lamp is (Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$). (in $W$)

The momentum of a photon of light of frequency $f$ is . . . . . . .

The number of photons per second on an average emitted by a source of monochromatic light of wavelength $600 \, nm$,when it delivers a power of $3.3 \times 10^{-3} \, W$,is: $(h = 6.6 \times 10^{-34} \, Js)$