Animated Solution for Physics - Dual Nature of Matter and Radiation: Ultraviolet light of wavelengths $800\text{ A˚}$ and $700\text{ A˚}$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy $1.8\text{ eV}$ and $4.0\text{ eV}$ respectively. Find the value of Planck's constant.

Imagine a solitary hydrogen atom, its single electron orbiting peacefully in the ground state. This is a state of deep stability, bound to the nucleus by an invisible electromagnetic tether. Suddenly, an invisible bullet of energy—an ultraviolet photon—strikes the atom. This isn't a gentle nudge; it's a catastrophic transfer of energy. The photon is completely absorbed, vanishing from existence, and its energy is transferred entirely to the electron.

If the photon packs enough punch, it shatters the electromagnetic tether, ripping the electron away from the nucleus. But the story doesn't end there. Any energy left over after paying the "ransom" (the ionization energy) is pocketed by the electron as kinetic energy, sending it flying off into the void. This beautiful, violent process is the photoelectric effect, and it is governed by a simple yet profound law of energy conservation formulated by Albert Einstein.

Einstein's photoelectric equation states that the energy of the incoming photon ($E_{\text{photon}}$) equals the work function ($\Phi$) plus the maximum kinetic energy of the ejected electron ($K$).

For a hydrogen atom in its ground state, the "work function" is simply its ionization energy, which we know from the Bohr model to be exactly $13.6\text{ eV}$. The energy of the photon can be expressed in terms of its wavelength $\lambda$ as $\frac{hc}{\lambda }$. Thus, our master equation becomes:

The problem provides us with two distinct scenarios. Let's focus on the first one, where a photon with a wavelength of $800\text{ A˚}$ ejects an electron with $1.8\text{ eV}$ of kinetic energy. We can plug these values directly into our master equation:

Adding the energies on the right side, we find that the total energy of the incident photon is $15.4\text{ eV}$. However, to find Planck's constant ($h$) in standard SI units (Joule-seconds), we must convert this energy from electron volts to Joules. We do this by multiplying by the elementary charge, $1.6\times 10^{-19}\text{ C}$:

Now, the path is clear. We isolate $h$ by multiplying both sides by the wavelength and dividing by the speed of light ($c=3\times 10^8\text{ m/s}$):

Rounding to two significant figures, we arrive at $6.6\times 10^{-34}\text{ J-s}$, which beautifully aligns with the accepted value of Planck's constant!

The Subtraction Method

You might be wondering, "Why did the problem give us a second wavelength of $700\text{ A˚}$?" While we could simply repeat the calculation above to verify our answer, there is a much more elegant way to use this extra data.

What if we didn't know the ionization energy of hydrogen? By setting up the photoelectric equation for both scenarios, we get a system of two equations:

If we subtract the first equation from the second, the unknown work function $\Phi$ completely cancels out!

This elegant subtraction method allows experimentalists to determine Planck's constant without needing prior knowledge of the material's binding energy. It's a testament to the power of algebra in unlocking the secrets of the quantum world.