Photoelectric effect experiments are performed using three different metal plates $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ having work functions $\phi_p=2.0 \mathrm{eV}, \phi_q=2.5 \mathrm{eV}$ and $\phi_r=3.0 \mathrm{eV}$, respectively. A light beam containing wavelengths of $550 \mathrm{~nm}, 450 \mathrm{~nm}$ and $350 \mathrm{~nm}$ with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is [Take $h c=1240 \mathrm{eV} \mathrm{nm}$ ]

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.

$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power, emitting radiowaves of wavelength $500\; m$.

$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can percetve $(-10^{-10}\; W m ^{-2}$). Take the area of the pupil to be about $0.4 \;cm ^{2}$, and the average frequency of white light to be about $6 \times 10^{14}\; Hz$

There are ${n_1}$ photons of frequency ${\gamma _1}$ in a beam of light. In an equally energetic beam, there are ${n_2}$ photons of frequency ${\gamma _2}$. Then the correct relation is

What is velocity of photon ? 

The momentum of the photon of wavelength $5000\,\mathring A$ will be

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 $.........\times 10^{-17} kg\,m / s$