Q.Threshold frequency, ν0 is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency 1.0×10^15 s^-1 was allowed to hit a metal surface, an electron having 1.988 × 10^-19 J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

Concept & Intuition

A unit cell is the smallest repeating unit of a crystal lattice. If we know the edge length of the cubic cell, we can compute its volume. The total volume of the metal is given, and since the metal is entirely made up of these unit cells packed together, the number of unit cells is simply the total volume divided by the volume of one cell. No need to worry about atoms per cell or packing fraction — the question asks only for the number of unit cells, not atoms.

1. Find the volume of one unit cell The edge length is given as $4\ \text{Å}$ (angstroms). Convert to cm: $1\ \text{Å} = 10^{-8}\ \text{cm}$, so

Volume of a cubic cell:

2. Divide total volume by cell volume Total volume given: $V_{\text{total}} = 128\ \text{cm}^3$. Number of unit cells:

Simplify:

The core idea here is to connect the microscopic property of kinetic energy of gas molecules to the macroscopic properties of temperature and pressure. For an ideal gas, the average translational kinetic energy of its molecules is directly proportional to the absolute temperature. Once we determine the temperature from the given kinetic energy, we can use the ideal gas law to find the pressure.

1. Determine the absolute temperature ($T$) from the kinetic energy:

   For one mole of an ideal gas, the average translational kinetic energy ($K_E$) is given by the formula:

   $K_E = \frac{3}{2}RT$

   Here, $R$ is the molar gas constant. Since the kinetic energy is given in Joules per mole, we use $R = 8.314~\text{J mol}^{-1}~\text{K}^{-1}$.

   We are given $K_E = 3000~\text{J mol}^{-1}$. Substituting this value:

   $3000~\text{J mol}^{-1} = \frac{3}{2} \times (8.314~\text{J mol}^{-1}~\text{K}^{-1}) \times T$

   To find $T$:

   $T = \frac{2 \times 3000~\text{J mol}^{-1}}{3 \times 8.314~\text{J mol}^{-1}~\text{K}^{-1}}$

   $T = \frac{6000}{24.942}~\text{K}$

   $T \approx 240.55~\text{K}$

2. Calculate the pressure ($P$) using the Ideal Gas Law:

   Now that we have the temperature, we can use the ideal gas law, which relates pressure, volume, number of moles, and temperature:

   $PV = nRT$

   We are given:

   • Number of moles, $n = 3~\text{mol}$ …

The number average molecular weight ($M_n$) is like the ordinary average you’d compute in a classroom: add up everyone’s weight, then divide by the number of people. For a polymer, it’s the total mass of all chains divided by the total number of chains. This is the simplest measure of “average” chain size, and it’s especially sensitive to the number of small molecules present.

Let’s work it out step by step.

1. Find the total number of molecules. We have 800 molecules of mass 1000, 100 of mass 2000, and 100 of mass 5000. Total number:

2. Find the total mass of all molecules. Multiply each group’s count by its molar mass and sum:

Compute each term:

• $800 \times 1000 = 800{,}000$
• $100 \times 2000 = 200{,}000$
• $100 \times 5000 = 500{,}000$ Sum:

3. Divide total mass by total number of molecules. $$$$

Concept: An electromagnetic wave consists of mutually perpendicular electric ($E$) and magnetic ($B$) field components that oscillate in phase and propagate together. They therefore have:

• the same wavelength $\lambda$,
• the same frequency $\nu$,
• the same speed $c$ (they are one wave). …

Concept & Intuition

The de-Broglie wavelength of a particle is given by $\lambda = h/p$, where $p$ is its momentum. Here we are given kinetic energy $K$, not momentum directly. For a non‑relativistic particle (the energy is tiny, so this is safe), kinetic energy and momentum are related by $K = p^2/(2m)$. So we can solve for $p = \sqrt{2mK}$ and then substitute into the de‑Broglie formula. The trick is to keep careful track of units — the answer is requested in nanometres.

Step‑by‑step solution

1. Write down the de‑Broglie relation

where $h = 6.63 \times 10^{-34}\ \text{J·s}$ is Planck’s constant and $p$ is the electron’s momentum.

2. Relate momentum to kinetic energy For a non‑relativistic particle,

Here $m = 9.1 \times 10^{-31}\ \text{kg}$ and $K = 4.55 \times 10^{-25}\ \text{J}$.

3. Substitute into the wavelength formula

4. Plug in the numbers First compute the product inside the square root:

Multiply the coefficients: $2 \times 9.1 \times 4.55 = 2 \times 41.405 = 82.81$.

Multiply the powers of ten: $10^{-31} \times 10^{-25} = 10^{-56}$.

So

5. Take the square root

Since $\sqrt{8.281} \approx 2.878$ and $10^{-27.5} = 10^{-27} \times 10^{-0.5} = 10^{-27} \times 0.3162$, we get

(A cleaner way: $\sqrt{8.281 \times 10^{-55}} = \sqrt{82.81 \times 10^{-56}} = \sqrt{82.81} \times 10^{-28} = 9.1 \times 10^{-28}$ because $\sqrt{82.81} = 9.1$ exactly — nice numbers given!) …

The key idea is that the energy of an electron in a hydrogen atom depends only on the principal quantum number $n$, and it varies inversely as $n^2$. This comes from the Bohr model, where the electron’s total energy (kinetic + potential) is quantized. The negative sign means the electron is bound to the nucleus; the more negative the energy, the more tightly bound it is. So for $n=1$ (ground state), the energy is most negative; for higher $n$, the energy becomes less negative (closer to zero). The ratio of energies is simply the ratio of $1/n^2$ values.

1. Recall the formula The energy of an electron in the $n$-th orbit of a hydrogen atom is

The constant $13.6$ eV is the ground state energy (for $n=1$). This formula is derived from balancing Coulomb attraction and centripetal force, plus quantization of angular momentum.

2. Write the energies for $n=1,2,3$

3. Find the ratio Since the constant $-13.6$ eV cancels out, the ratio of the energies is

4. Eliminate fractions …

Concept & Intuition

Graham’s law tells us that lighter gases diffuse faster. Specifically, the rate of diffusion is inversely proportional to the square root of the molar mass:

So if gas A diffuses $\sqrt{5}$ times faster than gas B, it must be lighter — and the ratio of their molar masses follows directly from squaring the inverse ratio of rates. Many students mistakenly think “$\sqrt{5}$ times more” means $r_A = r_B + \sqrt{5}\,r_B$, but here “more than” simply means the ratio is $\sqrt{5}$ (i.e., $r_A / r_B = \sqrt{5}$). That’s the key.

Step-by-step solution

1. Write Graham’s law For two gases A and B at the same temperature and pressure:

where $r$ is the rate of diffusion and $M$ is the molar mass.

2. Plug in the given ratio We are told $r_A$ is $\sqrt{5}$ times $r_B$, so:

Substitute into Graham’s law:

3. Square both sides

This tells us gas B is 5 times heavier than gas A. …

Concept and Intuition

This problem is a direct application of Einstein’s photoelectric equation. When light hits a metal surface, electrons are ejected if the photon energy exceeds the work function. The stopping potential is the voltage that just stops the most energetic electrons, so $eV_s$ equals the maximum kinetic energy of those electrons. Therefore, the photon energy is split into two parts: overcoming the work function and giving the electron its kinetic energy. We can combine these to find the wavelength.

Step-by-step solution

1. Write Einstein’s photoelectric equation The maximum kinetic energy of an ejected electron is

where $h$ is Planck’s constant, $\nu$ is the frequency of the radiation, and $\phi$ is the work function.

2. Relate stopping potential to kinetic energy The stopping potential $V_s$ is the voltage that reduces the photocurrent to zero, meaning

where $e$ is the elementary charge. Here $V_s = 2.63$ V, so

(since $1$ eV is the energy gained by an electron through $1$ V potential difference).

3. Combine the equations

Given $\phi = 4.3$ eV and $e V_s = 2.63$ eV, the photon energy is

4. Convert photon energy to wavelength …

Concept and Intuition

This is a classic photoelectric effect problem. When light hits a metal surface, electrons are ejected if the photon energy exceeds the work function. The stopping potential $V_s$ is the voltage needed to stop the most energetic photoelectrons, so the maximum kinetic energy equals $eV_s$. The key equation is:

where $h\nu$ is the photon energy, $\phi$ is the work function, and $e$ is the elementary charge. We are given $V_s = 2.63~\text{V}$ and $\phi = 4.3~\text{eV}$. Since $eV_s$ is in electronvolts (because $e$ times volts gives eV), we can work entirely in eV and then convert to wavelength.

Step-by-step solution

1. Write the photoelectric equation in eV units The maximum kinetic energy of ejected electrons is $K_{\text{max}} = eV_s$. In eV units, $e = 1$ (since 1 electronvolt is the energy gained by an electron through 1 volt), so:

The equation becomes:

2. Convert photon energy to wavelength The relation between photon energy and wavelength is:

A useful constant: $hc = 1240~\text{eV·nm}$ (approximately). This is derived from $h = 4.135667 \times 10^{-15}~\text{eV·s}$ and $c = 2.998 \times 10^8~\text{m/s}$, giving $hc \approx 1240~\text{eV·nm}$).

So:

$\lambda = \frac{hc}{E} = \frac{1240~\text{eV·nm}}{6.93~\text{eV}}$ …

The core idea here is the photoelectric effect — Einstein’s revolutionary insight that light behaves as packets of energy (photons), each carrying energy $h\nu$. When a photon strikes a metal surface, it transfers its energy to an electron. Part of that energy goes into overcoming the work function $\phi$ (the minimum energy needed to free the electron), and the rest becomes the electron’s kinetic energy.

The stopping potential $V_0$ is the voltage that just barely stops the most energetic photoelectrons from reaching the other electrode. At that point, the electrical work $eV_0$ done on the electron equals its maximum kinetic energy. So the physics is: photon energy in = work function out + maximum kinetic energy out. That’s the equation we need.

1. Start with Einstein’s photoelectric equation for the maximum kinetic energy of an emitted electron:

Here $h$ is Planck’s constant, $\nu$ is the frequency of incident light, and $\phi$ is the work function of the metal.

2. The stopping potential $V_0$ is defined such that the electrical work done to stop the electron equals its maximum kinetic energy:

where $e$ is the elementary charge.

3. Substitute $K_{\text{max}}$ from step 1 into step 2:

4. Solve for $V_0$ by dividing both sides by $e$: …