In a photoemissive cell with executing wavelength $\lambda $, the fastest electron has speed $v.$ If the exciting wavelength is changed to $\frac{{3\lambda }}{4}$, the speed of the fastest emitted electron will be

Kinetic energy with which the electrons are emitted from the metal surface due to photoelectric effect is

A small object at rest, absorbs a light pulse of power $20\,mW$ and duration $300\,ns$. Assuming speed of light as $3 \times 10^8\,m / s$. the momentum of the object becomes equal to $………\times 10^{-17} kg\,m / s$

A convex lens of focal length $40 \mathrm{~cm}$ forms an image of an extended source of light on a photoelectric cell. A current I is produced. The lens is replaced by another convex lens having the same diameter but focal length $20 \mathrm{~cm}$. The photoelectric current now is:

A $2\,mW$ laser operates at a wavelength of $500\,nm.$ The number of photons that will be emitted per second is [Given Planck’s constant $h = 6.6 \times 10^{-34}\,Js,$ speed of light $c = 3.0\times 10^8\,m/s$ ]