Q.A man wants to travel from corner A of a square field to the diagonally opposite corner C. The square has side $100\ \text{m}$. A smaller square of side $50\ \text{m}$ sits at the centre of the field (so its edges are $25\ \text{m}$ from each side of the big square) and is filled with sand; its diagonal lies along the diagonal AC. Outside this central sand square the man walks at $1\ \text{m/s}$; inside the sand he can walk only at speed $v\ \text{m/s}$, with $v < 1$. Find the smallest value of $v$ for which travelling along the straight diagonal path from A to C (which cuts through the sand) is faster than the quickest path that stays entirely outside the sand.

The core idea here is the photoelectric effect equation from Einstein: the energy of an incident photon ($h\nu$) is used first to overcome the work function ($h\nu_0$) and the remainder becomes the kinetic energy of the emitted electron. That kinetic energy is $\frac12 m v^2$, so equating and solving for $v$ gives the velocity.

1. Write the photoelectric equation The maximum kinetic energy of an emitted photoelectron is

Here $\nu$ is the frequency of incident light and $\nu_0$ is the threshold frequency (the minimum frequency needed to eject an electron).

2. Relate kinetic energy to velocity For a non‑relativistic electron (which is the case here),

where $m$ is the electron’s mass and $v$ is its speed.

3. Equate and solve for $v$ Set the two expressions equal:

Multiply both sides by 2:

Divide by $m$:

Take the square root (velocity is positive):

4. Match with the options …

The key concept here is the photoelectric effect, where light of frequency $f$ ejects electrons from a metal surface. The maximum kinetic energy $K_{\text{max}}$ of the emitted electrons is given by Einstein’s equation:

where $h$ is Planck’s constant, $f$ is the incident frequency, and $f_0$ is the threshold frequency (the minimum frequency needed to eject electrons). The term $hf_0$ is the work function of the metal.

Why this works: The incident photon gives energy $hf$ to an electron. Some of that energy is used to overcome the binding energy (work function $hf_0$), and the rest appears as kinetic energy. So, if we know $K_{\text{max}}$ and $f$, we can solve for $f_0$ by rearranging.

Let’s go step by step.

1. Write the photoelectric equation

We want $f_0$, so isolate it:

2. Insert the given values

   • $f = 1.0 \times 10^{15}$ Hz
   • $K_{\text{max}} = 2 \times 10^{-19}$ J
   • $h = 6.6 \times 10^{-34}$ Js

   Compute $\frac{K_{\text{max}}}{h}$:

3. Subtract to find $f_0$ $$$$

The question gives you drift speed $v_d$, resistivity $\rho$, electron density $n$, and asks for the mean free path $\lambda$. The link between these quantities runs through the relaxation time $\tau$ — the average time between collisions of an electron. Once you have $\tau$, the mean free path is simply $\lambda = v_d \tau$ (since drift speed is the average speed between collisions in the direction of the field).

But you don't have $\tau$ directly. You have two separate equations that each involve $\tau$: one from the definition of drift speed, and one from the definition of resistivity. Equating them lets you eliminate $\tau$ and solve for $\lambda$ in one clean step.

1. Drift speed relation. In an electric field $E$, an electron accelerates for a time $\tau$ between collisions, gaining a drift speed

This is just $v = a\tau$ with $a = eE/m$.

2. Resistivity relation. Resistivity is related to the relaxation time by

This comes from the Drude model: conductivity $\sigma = \frac{ne^2\tau}{m}$, and $\rho = 1/\sigma$.

3. Eliminate $E$ and $\tau$. From the drift equation, $\tau = \frac{m v_d}{eE}$. From the resistivity equation, $\tau = \frac{m}{n e^2 \rho}$. Equate them:

Cancel $m$ and one $e$:

So $E = n e \rho v_d$. You don't actually need $E$ — but notice that the two $\tau$ expressions are equal, so you can directly write

That's a consistency check. The real shortcut: since $\lambda = v_d \tau$, substitute $\tau$ from the resistivity formula:

4. Plug in the numbers.

First compute $n e^2 \rho$:

$n e^2 \rho = (9 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times (1 \times 10^{-8}).$ …

The problem asks us to find the edge length of a KBr unit cell, given its density and molecular weight, and knowing it has a rock salt type structure. This is a classic application of the density formula for crystalline solids.

The core idea is that the density of a unit cell is determined by the total mass of the atoms (or ions) within that unit cell divided by the volume of the unit cell.

Concept and Intuition

1. Rock Salt Structure: KBr crystallizes in a rock salt structure, which is a face-centered cubic (FCC) lattice. In this arrangement, one type of ion (say, Br⁻) forms an FCC lattice, and the other type of ion (K⁺) occupies all the octahedral voids.

   • For an FCC lattice, there are 4 atoms (or ions) effectively present per unit cell.
   • For a rock salt structure like KBr, this means there are 4 K⁺ ions and 4 Br⁻ ions per unit cell. Therefore, the number of formula units of KBr per unit cell, denoted by $Z$, is 4.

2. Density Formula: The density ($\rho$) of a unit cell is given by the formula:

   $\rho = \frac{Z \cdot M}{a^3 \cdot N_A}$

   Where:

   • $\rho$ is the density of the crystal.
   • $Z$ is the number of formula units per unit cell.
   • $M$ is the molecular weight (or molar mass) of the compound.
   • $a$ is the edge length of the unit cell.
   • $N_A$ is Avogadro's number ($6.022 \times 10^{23}\ \text{mol}^{-1}$).

We are given $\rho$, $M$, and we've determined $Z$. We need to find $a$. We can rearrange the formula to solve for $a^3$ and then take the cube root.

Step-by-Step Solution

1. Identify the given values and constants:

   • Density, $\rho = 3.70\ \text{g/cm}^3$
   • Molecular weight of KBr, $M = 120\ \text{g/mol}$
   • Avogadro's number, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$

2. Determine the number of formula units per unit cell ($Z$):

   As KBr has a rock salt type structure, it is an FCC arrangement of ions. In an FCC unit cell, there are 4 formula units.

   Therefore, $Z = 4$.

3. Rearrange the density formula to solve for $a^3$:

   From the formula $\rho = \frac{Z \cdot M}{a^3 \cdot N_A}$, we can isolate $a^3$:

   $a^3 = \frac{Z \cdot M}{\rho \cdot N_A}$

4. Substitute the values into the rearranged formula:

   $a^3 = \frac{4 \cdot 120\ \text{g/mol}}{3.70\ \text{g/cm}^3 \cdot 6.022 \times 10^{23}\ \text{mol}^{-1}}$ …

Ray optics (geometrical optics) treats light as straight rays, ignoring diffraction. But light is a wave, so diffraction always spreads a beam. The question asks: for a given aperture width $a$ and wavelength $\lambda$, at what distance does the spreading due to diffraction become negligible compared to the aperture size? That distance is called the Fresnel distance (or Rayleigh distance) $Z_F = a^2/\lambda$. Beyond this, the beam’s angular spread $\theta \approx \lambda/a$ causes the beam width to increase by roughly the original aperture size, so ray optics is a good approximation. Inside this distance, diffraction effects (like Fresnel fringes) are significant.

Let’s compute it step by step.

1. Identify the given quantities

   Aperture width: $a = 5$ mm = $5 \times 10^{-3}$ m

   Wavelength: $\lambda = 450$ nm = $450 \times 10^{-9}$ m = $4.5 \times 10^{-7}$ m

2. Recall the Fresnel distance formula

   The condition for ray optics to be valid is that the distance $L$ satisfies $L \gg a^2/\lambda$. The threshold distance is

This comes from comparing the path difference at the edge of the aperture to $\lambda/2$ — when the Fresnel number $N = a^2/(\lambda L) \approx 1$, diffraction is significant; for $N \ll 1$, ray optics works.

3. Plug in the numbers

Simplify: