An electron and proton are separated by a large distance. The electron starts approaching the proton

Q: Why is the initial potential energy of the electron considered zero?

When the electron is separated by a 'large distance' (effectively infinity) from the proton, the electrostatic potential energy $U = \frac{kq_1q_2}{r}$ approaches zero. Thus, its total initial energy is just its kinetic energy.

Q: Why does 'second excited state' mean $n=3$ and not $n=2$?

The lowest energy level (ground state) is $n=1$. The first time the electron gets excited, it goes to $n=2$ (first excited state). Therefore, the second time it gets excited, it reaches $n=3$ (second excited state).

Q: Why do we subtract a negative energy when calculating the photon energy?

The photon carries away the *loss* in energy. The change in energy is $E_{initial} - E_{final}$. Since bound states have negative energy, subtracting a negative value mathematically adds to the total energy released, which makes perfect physical sense!

Q: Can I use $hc = 1242 \text{ eV}\cdot\text{nm}$ instead of $12400 \text{ eV}\cdot\text{\AA}$?

Absolutely! Both are valid approximations. Since the threshold wavelength is given in Angstroms ($4000 \text{ \AA}$), using $12400 \text{ eV}\cdot\text{\AA}$ makes the calculation slightly faster and cleaner without needing unit conversions.

Q: What if the incident photon energy was less than the work function?

If $E_\gamma < \phi$, the photon wouldn't have enough energy to break the electron free from the metal's surface. In that case, no photoelectron would be emitted, and the kinetic energy would be zero.

Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 \AA. What is the maximum kinetic energy of the emitted photoelectron?

Options:

(A) 7.61 eV

(B) 1.41 eV

(C) 3.3 eV

(D) No photoelectron would be emitted

Sigma Percentile · Accepted Answer

The problem we are tackling today is a beautiful symphony of two of the most profound concepts in modern physics: the Bohr Model of the atom and Einstein's Photoelectric Effect. It is a journey that takes us from the vast emptiness of space, into the heart of a hydrogen atom, and finally to the surface of a metal plate.

Let's dive into this cosmic dance!

### The Cosmic Dance: Electron Meets Proton

Imagine an electron, wandering through the vacuum of space, separated by a massive distance from a lonely proton. Because they are so far apart, the electrostatic potential energy between them is effectively zero. However, this electron is not sitting still; it is hurtling towards the proton with an initial kinetic energy of $3 	ext{ eV}$.

Therefore, the total initial energy of our free electron is simply its kinetic energy:

$$E_i = 3 	ext{ eV}$$

As the electron gets closer, the electrostatic pull of the proton becomes irresistible. The proton captures the electron, giving birth to a brand new hydrogen atom! But the electron doesn't just crash into the nucleus; it settles into a specific quantum orbit. The problem tells us it lands in the **second excited state**.

We must be careful here. The ground state is $n=1$, and the first excited state is $n=2$. Thus, the second excited state corresponds to the principal quantum number **$n=3$**.

### The Energy Drop and Photon Birth

Now that the electron is bound to the proton, it has a new energy level. According to the Bohr model, the energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is given by the famous formula:

$$E_n = \frac{-13.6}{n^2} 	ext{ eV}$$

Let's plug in our value of $n=3$ to find the final energy of the electron:

$$E_f = \frac{-13.6}{3^2} = \frac{-13.6}{9} = -1.51 	ext{ eV}$$

Notice what just happened. The electron started with a positive energy of $3 	ext{ eV}$ and ended up with a negative energy of $-1.51 	ext{ eV}$. The universe demands that energy be conserved. So, where did this lost energy go?

It is released into the universe as a brilliant flash of light—a photon! The energy of this emitted photon is exactly equal to the difference between the initial and final energies of the electron:

$$E_\gamma = E_i - E_f$$

Substituting our values, we get:

$$E_\gamma = 3 - (-1.51) = 4.51 	ext{ eV}$$

This single photon, carrying $4.51 	ext{ eV}$ of energy, now speeds away from the newly formed atom.

### The Photoelectric Finale

Our photon's journey is not over. It travels until it violently strikes a photosensitive metal surface. We are told this metal has a threshold wavelength ($\lambda_0$) of $4000 	ext{ \AA}$.

Before we can determine if an electron is ejected from this metal, we need to know the metal's **work function** ($\phi$). The work function is the minimum energy required to rip an electron from the metal's surface, and it is directly related to the threshold wavelength:

$$\phi = \frac{hc}{\lambda_0}$$

To save time and avoid messy unit conversions, we use the incredibly handy approximation $hc \approx 12400 	ext{ eV}\cdot	ext{\AA}$. Let's calculate the work function:

$$\phi = \frac{12400 	ext{ eV}\cdot	ext{\AA}}{4000 	ext{ \AA}} = 3.1 	ext{ eV}$$

Now for the grand finale. We apply **Einstein's Photoelectric Equation**, which states that the maximum kinetic energy of an emitted photoelectron is the energy of the incoming photon minus the work function of the metal:

$$K_{max} = E_\gamma - \phi$$

We have all the pieces of the puzzle. Let's substitute them in:

$$K_{max} = 4.51 	ext{ eV} - 3.1 	ext{ eV} = 1.41 	ext{ eV}$$

And there we have it! The maximum kinetic energy of the ejected photoelectron is **$1.41 	ext{ eV}$**. This perfectly matches option (b).

This problem is a fantastic reminder of how interconnected physics truly is. We tracked a single packet of energy as it transformed from the kinetic energy of a free particle, to the binding energy of an atom, to the electromagnetic energy of a photon, and finally back to the kinetic energy of a new photoelectron. Truly elegant!

Select Answer:

Visualized Solution

$K_{i}$ and $U_{i}$ of Free Electron

Electron Capture State

Energy in $n^{th}$ Orbit

Final Energy Calculation

Energy of Emitted Photon

Calculating $E_{γ}$

Photon Strikes Metal

Work Function Formula

Calculating Work Function

Einstein's Photoelectric Equation

Final Kinetic Energy

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Cosmic Dance

The Energy Drop and Photon Birth

The Photoelectric Finale

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