Q.The diameter of zinc atom is 2.6 Å. Calculate

The problem connects two fundamental ideas from early quantum theory: the Bohr model of the hydrogen atom and de Broglie’s wave-particle duality. In the Bohr model, an electron in a stable orbit has an angular momentum that is an integer multiple of $\hbar$. De Broglie later suggested that an electron behaves like a standing wave around the nucleus; for the wave to be in phase after one full trip, the circumference of the orbit must be an integer number of wavelengths. This condition is exactly equivalent to Bohr’s quantization rule, and it gives us a direct way to relate the orbit’s radius to the electron’s wavelength.

Let’s work through it step by step.

1. Recall the de Broglie standing wave condition For an electron in a circular orbit of radius $r$, the circumference is $2\pi r$. For a stable standing wave, this circumference must equal an integer multiple of the de Broglie wavelength $\lambda$:

Here $n$ is the principal quantum number (the orbit number).

2. Identify the given wavelength The problem states the wavelength in orbit X is $6\pi a_0$, where $a_0$ is the Bohr radius (radius of the first orbit, $n=1$). So:

3. Express the radius of the $n$-th Bohr orbit In the Bohr model, the radius of the $n$-th orbit is:

This is a standard result: the radius scales as $n^2$.

4. Apply the standing wave condition to orbit X …

The key idea is that density in a crystal relates the mass of atoms in a unit cell to the volume of that cell. For a face-centred cubic (FCC) lattice, each unit cell contains 4 atoms ($Z = 4$). We can rearrange the density formula to isolate molar mass $M$.

Step-by-step reasoning:

1. Recall the density formula for a crystal The density $\rho$ of a crystalline solid is given by:

where:

• $Z$ = number of atoms per unit cell,
• $M$ = molar mass (kg/mol),
• $N_A$ = Avogadro’s number (mol$^{-1}$),
• $a$ = edge length of the unit cell (m).

2. Identify the given values

   • $\rho = 2.7 \times 10^3 \ \mathrm{kg \ m^{-3}}$
   • $a = 405 \ \mathrm{pm} = 405 \times 10^{-12} \ \mathrm{m} = 4.05 \times 10^{-10} \ \mathrm{m}$
   • $N_A = 6.0 \times 10^{23} \ \mathrm{mol^{-1}}$
   • For FCC, $Z = 4$ (atoms at corners contribute 1/8 each, face centres contribute 1/2 each: $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$).

3. Rearrange the formula to solve for $M$

4. Compute $a^3$

(Since $4.05^3 = 4.05 \times 4.05 \times 4.05 = 16.4025 \times 4.05 \approx 66.43$.)

5. Plug in the numbers

First, multiply the constants:

Then combine powers of ten:

So:

6. Simplify …

The key idea here is that in the Bohr model of the hydrogen atom, the angular momentum of the electron is quantized in units of $h/2\pi$. The given angular momentum value directly tells you the principal quantum number $n$. Once you know $n$, the de Broglie wavelength follows from the standing-wave condition: the circumference of the orbit must be an integer multiple of the wavelength.

Let’s work through it step by step.

1. Find the quantum number $n$ from the angular momentum. In the Bohr model, the angular momentum $L$ of the electron in the $n$th orbit is

You are given $L = 1.051 \times 10^{-34}$ Js, $h = 6.6 \times 10^{-34}$ Js, and $\pi = 3.14$.

Compute $h/2\pi$:

So $L = 1.051 \times 10^{-34}$ Js is essentially equal to $1.05 \times 10^{-34}$ Js, meaning $n = 1$.

2. Recall the de Broglie wavelength condition in the Bohr model. For a stable orbit, the electron’s de Broglie wave must form a standing wave around the circumference:

where $r$ is the radius of the $n$th orbit. So

3. Find the radius for $n=1$. The Bohr radius for $n=1$ is

(You should know this standard value; it comes from $r_n = n^2 \times 0.529$ Å.)

4. Calculate the wavelength. …

In an FCC (face-centered cubic) unit cell, octahedral voids are located at two key positions: at the body center (one per cell) and at the midpoints of each edge (12 edges, each shared by 4 cells, giving 12/4 = 3 per cell). So there are 4 octahedral voids per FCC unit cell in total.

The question asks for the distance between two octahedral voids. The nearest pair of octahedral voids in FCC are the one at the body center and the one at the midpoint of an edge. Let’s see why.

The body-centered octahedral void is at coordinates $(\frac12, \frac12, \frac12)$. An edge-centered octahedral void lies at, say, $(\frac12, 0, 0)$ (midpoint of an edge along the x-axis). The distance between these two points is the shortest separation between any two octahedral voids in the FCC lattice.

1. Relate the body diagonal to the edge length. The body diagonal of a cube of edge length $a$ is $a\sqrt{3}$. We are told this length is $x$ Å. So:

2. Find the coordinates of the two octahedral voids.

   Take the body-centered void: $(\frac12, \frac12, \frac12)$.

   Take an edge-centered void on the x-axis edge at y=0, z=0: $(\frac12, 0, 0)$.

   (Any edge-centered void adjacent to the body center will give the same distance by symmetry.)

3. Compute the distance between them.

   The difference in coordinates:

So the distance in terms of $a$ is: …

The core idea here is Heisenberg’s uncertainty principle, which states that the product of the uncertainties in position ($\Delta x$) and momentum ($\Delta p$) has a minimum value set by Planck’s constant. The given uncertainty in velocity is a cleverly disguised form of the uncertainty in momentum, since momentum $p = mv$. Once you connect the given expression to $\Delta p$, the ratio becomes straightforward.

Let’s work through it step by step.

1. Relate the given uncertainty to momentum The uncertainty in velocity is given as $\Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$. Since momentum $p = mv$, the uncertainty in momentum is $\Delta p = m \Delta v$. Substituting:

2. Apply Heisenberg’s uncertainty principle The principle states:

For the minimum uncertainty (which is what the problem implies), we take the equality:

3. Find $\Delta x$ Substitute $\Delta p$ from step 1:

Solve for $\Delta x$:

4. Compute the ratio $\Delta x : \Delta p$ …

The question describes a core limitation in quantum mechanics: you cannot know exactly where an electron is and exactly how fast it is moving at the same time. This isn't a problem with our instruments—it's a fundamental law of nature. Let's see why each option fits or doesn't.

1. Identify the core concept.

   The statement "both the position and exact velocity of an electron in an atom cannot be determined simultaneously and accurately" is the textbook definition of the Heisenberg uncertainty principle. Formally, it states that the product of the uncertainty in position ($\Delta x$) and the uncertainty in momentum ($\Delta p$) is at least a constant: $\Delta x \cdot \Delta p \ge \frac{\hbar}{2}$. Since velocity is directly related to momentum ($p = mv$), this principle applies directly to the statement.

2. Eliminate the other options.

   • (A) de Broglie principle: This principle says that every moving particle has a wavelength associated with it ($\lambda = h/p$). It explains wave-particle duality, not the limitation on simultaneous measurement.
   • (B) Hamiltonian law: This refers to the Hamiltonian formulation of classical mechanics (or the Hamiltonian operator in quantum mechanics), which describes the total energy of a system. It has nothing to do with measurement uncertainty.
   • (D) Bohr theory of hydrogen atom: Bohr's model assumes electrons orbit the nucleus in fixed, well-defined paths (like planets). In that model, both position and velocity are simultaneously knowable—which is exactly the opposite of the given statement. Bohr's theory was later superseded by quantum mechanics precisely because it violated the uncertainty principle. …

The question asks why diamond is the hardest natural substance while graphite is soft enough to be used as a pencil lead. Both are pure carbon — so the difference cannot be about the element itself. It must be about how the carbon atoms are arranged and bonded.

The key concept here is allotropy: different structural forms of the same element. Diamond and graphite are allotropes of carbon. Their properties — hardness, electrical conductivity, lubricating ability — are direct consequences of their bonding and geometry.

Let’s walk through the reasoning step by step.

1. Bonding in diamond

   Each carbon atom in diamond forms four covalent bonds with four neighbouring carbon atoms. These bonds are arranged in a tetrahedral geometry (bond angle $109.5^\circ$). This creates a three-dimensional network of strong covalent bonds extending in all directions. There are no weak layers or planes — every atom is locked in place. To break or deform diamond, you would have to break many strong covalent bonds simultaneously. That is why diamond is the hardest known natural material.

2. Bonding in graphite

   In graphite, each carbon atom forms three covalent bonds with three neighbours, giving a hexagonal planar sheet (like chicken wire). The fourth valence electron is delocalised, moving freely between the sheets. The sheets themselves are held together by weak van der Waals forces, not covalent bonds. These forces are easy to overcome — the sheets can slide past one another with little effort. That is why graphite feels slippery and soft, and why it is used as a lubricant and in pencil leads.

3. Comparing the number of bonded neighbours …

The Heisenberg Uncertainty Principle is a fundamental concept in quantum mechanics that describes a limit to the precision with which certain pairs of physical properties of a particle, such as position and momentum, can be known simultaneously. It's not a limitation of our measuring instruments, but a fundamental property of nature itself.

The core idea is that the act of measuring one property inevitably disturbs the other. For instance, to measure the position of an electron, you might shine light on it. The photons of light carry energy and momentum, and when they collide with the electron, they change its momentum in an unpredictable way. Conversely, if you try to precisely measure its momentum, you lose information about its exact position.

The significance of this principle depends on the scale of the objects involved. The effect is only noticeable when the product of the uncertainties in position and momentum is comparable to Planck's constant, $h$, which is an extremely small number.

1. State the Heisenberg Uncertainty Principle: The principle is mathematically expressed as:

Here, $ \Delta x $ represents the uncertainty in the particle's position, and $ \Delta p $ represents the uncertainty in its momentum. The term $h$ is Planck's constant, and its value is approximately $6.626 \times 10^{-34}\,\mathrm{J\,s}$.

> [!FORMULA]
> The Heisenberg Uncertainty Principle: $ \Delta x \Delta p \ge \frac{h}{4\pi} $

2. Analyze the implications for macroscopic objects:

Consider macroscopic objects like planets, cricket balls, or cars. These objects have relatively large masses. Let's take a cricket ball with a mass of $500\,\mathrm{g}$ ($0.5\,\mathrm{kg}$). Even if we could measure its position with an incredibly high precision, say, an uncertainty of $\Delta x = 1\,\mathrm{nm}$ ($10^{-9}\,\mathrm{m}$), the minimum uncertainty in its momentum would be:

Since $ \Delta p = m \Delta v $, the uncertainty in its velocity would be:

This uncertainty in velocity is astronomically small and completely unobservable in practice. For all practical purposes, we can consider both the position and momentum of a cricket ball (or a planet or a car) to be known simultaneously with arbitrary precision. Therefore, the uncertainty principle is not significant for macroscopic objects.

3. Analyze the implications for microparticles:

Now, consider a microparticle like an electron, which has a mass of approximately $9.1 \times 10^{-31}\,\mathrm{kg}$. If we try to measure its position with the same uncertainty of $\Delta x = 1\,\mathrm{nm}$, the minimum uncertainty in its momentum is still $\Delta p \approx 5.27 \times 10^{-26}\,\mathrm{kg\,m/s}$. However, the uncertainty in its velocity would be: …