Q.When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68 \times 10^{5}\ J\ mol^{-1}$. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?

The photoelectric effect is a beautiful demonstration of the particle nature of light. When a photon strikes a metal surface, it can transfer its entire energy to a single electron. But the electron doesn’t get to keep all that energy — it must first pay an “exit fee” (the work function) to escape the metal. What remains is the electron’s kinetic energy. So the emitted electron’s total energy (kinetic + any internal energy) is always smaller than the photon’s energy.

Let’s walk through the reasoning step by step.

1. Einstein’s photoelectric equation The energy of a single photon is $E_{\text{photon}} = hf$, where $h$ is Planck’s constant and $f$ is the frequency. When this photon is absorbed by an electron in the metal, the electron gains that energy. However, to leave the metal, the electron must overcome the work function $\phi$ (the minimum energy needed to escape). The remaining energy becomes the electron’s maximum kinetic energy:

This directly shows that the electron’s kinetic energy is less than the photon’s energy by exactly $\phi$.

2. What “energy of the emitted electron” means

   The question asks about the total energy of the emitted electron. For a free electron outside the metal, its total energy is just its kinetic energy (since potential energy is zero far away). So the emitted electron’s energy is $K_{\text{max}}$ (or less, if it loses energy in collisions inside the metal). In any case, it is always smaller than $hf$.

3. Why the other options are wrong

   • (A) “greater than the energy of the incident photon” — impossible, because energy is conserved; the electron cannot gain more than the photon gives. …

When a gas absorbs a single photon, it jumps to a higher energy level. It can then return to the ground state via two steps, emitting two photons whose energies add up to the energy of the absorbed photon. This is a classic application of energy conservation in atomic transitions:

Since energy is inversely proportional to wavelength ($E = hc/\lambda$), we can work directly with wavelengths — but carefully, because the relationship is not linear in wavelength.

1. Write the energy conservation equation Let $\lambda_a = 4.0 \times 10^{-7}\ \mathrm{m}$ be the absorbed wavelength, $\lambda_1 = 7.5 \times 10^{-7}\ \mathrm{m}$ be one emitted wavelength, and $\lambda_2$ be the unknown second emission wavelength. Using $E = \frac{hc}{\lambda}$, we have:

Cancel $hc$ (it’s nonzero):

2. Solve for $\lambda_2$ Rearranging:

Substitute the given values:

Compute each term:

Subtract:

Step 1 — Volume of the sample.

Step 2 — Volume of one unit cell (watch the unit conversion).

(using $1\ \text{pm}=10^{-10}\ \text{cm}$, since $1\ \text{cm}=10^{-2}\ \text{m}$ and $1\ \text{pm}=10^{-12}\ \text{m}$).

Step 3 — Number of unit cells in the sample.

Step 4 — Atoms per BCC unit cell ($Z$).

In a body-centred cubic cell:

• 8 corner atoms, each shared by 8 cells $\Rightarrow 8\times\frac18=1$ atom;
• 1 atom at the body centre, belonging entirely to that cell $\Rightarrow 1$ atom. $Z=1+1=2 .$ …