Consider a hydrogen-like ionised atom with atomic number Z with a single electron. In the emission s

Q: Why do we use $13.6 \text{ eV}$ in the formula?

Because $13.6 \text{ eV}$ is the ionization energy of a hydrogen atom (where $Z=1$). For a hydrogen-like atom with atomic number $Z$, the energy levels scale by a factor of $Z^2$.

Q: What does "hydrogen-like ionised atom" mean?

It refers to any atom or ion that has only one electron orbiting the nucleus, just like hydrogen. Examples include $\text{He}^+$, $\text{Li}^{2+}$, and $\text{Be}^{3+}$. Bohr's model strictly applies only to single-electron systems.

Q: Why is the energy of the $2 \rightarrow 1$ transition so much larger than the $3 \rightarrow 2$ transition?

The energy levels in Bohr's model are proportional to $-1/n^2$. The gap between $n=1$ and $n=2$ is the largest in the entire spectrum, accounting for $75\%$ of the ionization energy. As $n$ increases, the energy levels get closer together.

Q: Can $Z$ be a fraction or a negative number?

No, $Z$ represents the atomic number, which is the number of protons in the nucleus. It must be a positive integer. If your calculation gives a non-integer, you should recheck your math!

Q: What if the problem didn't mention the ionization energy of hydrogen?

It's a standard constant that you are expected to memorize for JEE. The ground state energy of hydrogen is $-13.6 \text{ eV}$, and its ionization energy is $+13.6 \text{ eV}$.

Question

Consider a hydrogen-like ionised atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom, the photon emitted in the $n = 2$ to $n = 1$ transition has energy $74.8	ext{ eV}$ higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionisation energy of the hydrogen atom is $13.6	ext{ eV}$. The value of $Z$ is ............ .

Sigma Percentile · Accepted Answer

### Unveiling the Unknown Atom

Imagine you are a detective, and your only clue is the light emitted by an unknown atom. This is exactly what we are doing in this problem! We are given a hydrogen-like ionised atom, meaning it has only one electron orbiting its nucleus, just like hydrogen. However, its atomic number $Z$ is unknown.

We are given a fascinating piece of information about its emission spectrum: the photon emitted when the electron jumps from the $n = 2$ orbit to the $n = 1$ orbit has an energy that is $74.8	ext{ eV}$ higher than the photon emitted during the $n = 3$ to $n = 2$ transition. Let's use Bohr's model to crack this case.

### The Master Equation

According to Bohr's model, the energy of a photon emitted during a transition from an initial state $n_i$ to a final state $n_f$ in a hydrogen-like atom is given by the formula:

$$ \Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV} $$

Here, $13.6	ext{ eV}$ is the ionization energy of hydrogen, and $Z$ is the atomic number of our mystery atom. Let's calculate the energy for the two specific transitions mentioned in the problem.

For the first transition ($n = 2 ightarrow n = 1$):

$$ \Delta E_{2 ightarrow 1} = 13.6 Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 Z^2 \left( 1 - \frac{1}{4} ight) = 13.6 Z^2 \left( \frac{3}{4} ight) $$

For the second transition ($n = 3 ightarrow n = 2$):

$$ \Delta E_{3 ightarrow 2} = 13.6 Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 Z^2 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 Z^2 \left( \frac{5}{36} ight) $$

### Setting Up the Clues

The problem states that the energy of the first transition is $74.8	ext{ eV}$ greater than the second. We can write this mathematically as:

$$ \Delta E_{2 ightarrow 1} = \Delta E_{3 ightarrow 2} + 74.8 $$

Now, let's substitute the expressions we found earlier into this equation:

$$ 13.6 Z^2 \left( \frac{3}{4} ight) = 13.6 Z^2 \left( \frac{5}{36} ight) + 74.8 $$

### The Final Calculation

To solve for $Z$, let's group the $Z^2$ terms on one side of the equation:

$$ 13.6 Z^2 \left( \frac{3}{4} - \frac{5}{36} ight) = 74.8 $$

Finding a common denominator of $36$ for the fractions inside the bracket:

$$ 13.6 Z^2 \left( \frac{27 - 5}{36} ight) = 74.8 $$

$$ 13.6 Z^2 \left( \frac{22}{36} ight) = 74.8 $$

We can simplify the fraction $\frac{22}{36}$ to $\frac{11}{18}$. Now, let's isolate $Z^2$:

$$ Z^2 = \frac{74.8 	imes 18}{13.6 	imes 11} $$

Notice a beautiful mathematical coincidence here! If you multiply $13.6$ by $11$, you get exactly $149.6$. And $149.6$ is exactly twice $74.8$!

$$ Z^2 = \frac{74.8}{149.6} 	imes 18 = \frac{1}{2} 	imes 18 = 9 $$

Taking the square root of both sides, we find:

$$ Z = 3 $$

**Conclusion:** The atomic number of our mystery atom is $3$. This means the hydrogen-like ion we were investigating is a doubly ionized Lithium atom ($	ext{Li}^{2+}$). We successfully identified the element using nothing but the light it emits!

Enter Numerical Value:

Visualized Solution

$Visualizing the Transitions$

$Bohr’s Energy Formula$

$Energy of First Transition$

$Energy of Second Transition$

$Applying the Given Condition$

$Solving for Z$

$Final Calculation$

$Conclusion$

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

Unveiling the Unknown Atom

The Master Equation

Setting Up the Clues

The Final Calculation

Similar Questions

A particular hydrogen like ion emits radiation of frequency $2.92 \times 1 0^{15}$ Hz when it makes transition from $n = 3$ to $n = 1$ . The frequency in Hz of radiation emitted in transition from $n = 2$ to $n = 1$ will be

If $13.6 eV$ energy is required to ionise the hydrogen atom, then the energy required to remove an electron from $n = 2$ is

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 ( $h c = 1242 eV nm$ )

The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Transitions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bohr’s Energy Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of First Transition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of Second Transition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Given Condition

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

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

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