A Hydrogen-like atom has atomic number Z. Photons emitted in the electronic transitions from level n

Q: Why do we use $13.6$ in the energy formula for a hydrogen-like atom?

$13.6 \text{ eV}$ is the Rydberg energy constant ($Rhc$) for hydrogen. For any hydrogen-like atom (single electron), the energy scales by the square of the nuclear charge, hence we multiply by $Z^2$.

Q: Can the atomic number $Z$ be a decimal?

Absolutely not. $Z$ represents the number of discrete protons in the nucleus. If your calculation yields $9.015$ for $Z^2$, you must round it to the nearest perfect square, which is $9$, making $Z=3$.

Q: Why did we add the work function and kinetic energy?

Think of the photon's energy as a budget. It must first 'pay' the work function to free the electron from the metal's surface. Whatever energy is left over becomes the kinetic energy of the freed electron. So, Total Energy = Work Function + Kinetic Energy.

Q: What does 'hydrogen-like atom' mean?

It means the atom or ion has only one electron orbiting the nucleus, just like hydrogen. Examples include $\text{He}^+$, $\text{Li}^{2+}$, and $\text{Be}^{3+}$. This allows us to use Bohr's model equations directly.

Q: Why is the energy of an orbit negative?

The negative sign indicates that the electron is bound to the nucleus. An energy of zero would mean the electron is completely free and at rest at an infinite distance from the nucleus.

Question

A Hydrogen-like atom has atomic number Z. Photons emitted in the electronic transitions from level n = 4 to level n = 3 in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is 1.95 eV. If the photoelectric threshold wavelength for the target metal is 310 nm, the value of Z is _______.
[Given: hc = 1240 eV-nm and Rhc = 13.6 eV, where R is the Rydberg constant, h is the Planck's constant and c is the speed of light in vacuum]

Sigma Percentile · Accepted Answer

### The Grand Intersection of Two Quantum Worlds

Imagine standing at the crossroads of two of the most revolutionary discoveries in modern physics: Bohr's atomic model and Einstein's photoelectric effect. This problem is a beautiful symphony that plays both tunes simultaneously. On one side of our laboratory, a mysterious hydrogen-like atom is undergoing an electronic transition. On the other side, a target metal sits quietly, waiting to be bombarded by the light emitted from this atom.

Our mission? To uncover the identity of this mysterious atom by finding its atomic number, $Z$. Let's break this grand experiment into two logical phases.

### Phase 1: Decoding the Target Metal

Before we look at the atom, let's analyze the target metal. We are told that the photoelectric threshold wavelength for this metal is $\lambda_{th} = 310 	ext{ nm}$.

What does this mean physically? The threshold wavelength corresponds to the exact minimum energy required to just pluck an electron out of the metal's surface, without giving it any extra kinetic energy. This minimum energy is called the **work function** ($W$).

Using the Planck-Einstein relation, we can calculate the work function:
$$W = \frac{hc}{\lambda_{th}}$$

We are generously given the value of $hc = 1240 	ext{ eV-nm}$. Let's substitute this in:
$$W = \frac{1240}{310} = 4 	ext{ eV}$$

So, the metal demands a toll of $4 	ext{ eV}$ from any incoming photon that wishes to eject an electron.

Now, the problem states that the ejected photoelectrons have a maximum kinetic energy of $K_{\max} = 1.95 	ext{ eV}$. According to **Einstein's Photoelectric Equation**, the total energy of the incoming photon ($E$) is split into two parts: paying the work function toll, and giving the electron its kinetic energy.
$$E = W + K_{\max}$$

Let's plug in our numbers:
$$E = 4 	ext{ eV} + 1.95 	ext{ eV} = 5.95 	ext{ eV}$$

This is a massive breakthrough! We now know that the photon striking the metal carries exactly $5.95 	ext{ eV}$ of energy. But where did this photon come from? It came from our mysterious atom.

### Phase 2: Unmasking the Mysterious Atom

Let's shift our focus to the hydrogen-like atom. The photon was born when an electron inside this atom jumped from the $n=4$ energy level down to the $n=3$ energy level.

According to Bohr's model, the energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by:
$$E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$$

The energy of the emitted photon is simply the difference in energy between these two levels:
$$\Delta E = E_4 - E_3$$

Let's calculate the energies of these individual levels. For $n=4$:
$$E_4 = -13.6 \frac{Z^2}{4^2} = -\frac{13.6}{16} Z^2 = -0.85 Z^2 	ext{ eV}$$

For $n=3$:
$$E_3 = -13.6 \frac{Z^2}{3^2} = -\frac{13.6}{9} Z^2 = -1.51 Z^2 	ext{ eV}$$

Now, let's find the energy difference:
$$\Delta E = -0.85 Z^2 - (-1.51 Z^2) = 0.66 Z^2 	ext{ eV}$$

This expression, $0.66 Z^2 	ext{ eV}$, represents the energy of the photon emitted by the atom.

### The Grand Finale: Equating the Energies

We have reached the climax of our problem. The energy of the photon emitted by the atom must perfectly match the energy of the photon that struck the metal.

Let's equate the two values we found:
$$0.66 Z^2 = 5.95$$

Now, it's just a matter of simple algebra to isolate $Z^2$:
$$Z^2 = \frac{5.95}{0.66} \approx 9.015$$

Since the atomic number $Z$ represents the number of protons in a nucleus, it must be a perfect integer. The closest perfect square is $9$.
$$Z = \sqrt{9} = 3$$

And there we have it! The atomic number is $3$. Our mysterious hydrogen-like atom is actually a doubly ionized Lithium atom ($	ext{Li}^{2+}$). The beauty of physics lies in how perfectly these different concepts interlock to reveal the hidden truths of the universe.

Enter Numerical Value:

Visualized Solution

$n = 4 \to n = 3 ⟹ K_{m a x}$

$W = \frac{h c}{λ _{t h}}$

$W = 4 eV$

$E = K_{m a x} + W$

$E = 5.95 eV$

$Δ E = 13.6 Z^{2} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$

$Δ E = 0.66 Z^{2} eV$

$0.66 Z^{2} = 5.95$

$Z = 3$

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

The Grand Intersection of Two Quantum Worlds

Phase 1

Phase 2

The Grand Finale

Similar Questions

Consider a hydrogen atom with its electron in the $n^{th}$ orbit. An electromagnetic radiation of wavelength $90 nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 eV$ , then the value of $n$ is ( $h c = 1242 eV nm$ )

$K_{α}$ wavelength emitted by an atom of atomic number $Z = 11$ is $λ$ . Find the atomic number for an atom that emits $K_{α}$ radiation with wavelength $4 λ$

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=4→n=3⟹Kmax​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=λth​hc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=4 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=Kmax​+W

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=5.95 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=13.6Z2(nf2​1​−ni2​1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=0.66Z2 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0.66Z2=5.95

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z=3