Q.Taking the Bohr radius as $a_0 = 53\ \text{pm}$, the radius of $\text{Li}^{++}$ ion in its ground state, on the basis of Bohr's model, will be about

For the Balmer series the lower level is $n=2$:

Given $\lambda=\dfrac{7.2}{R}$, so $\dfrac{1}{\lambda}=\dfrac{R}{7.2}$:

So $n=3$. The transition is between $n=3$ (higher) and $n=2$ (lower). Since orbital radius $r\propto n^2$:

Concept and Intuition

When an electron in a hydrogen atom drops from a higher energy level to a lower one, it emits a photon whose wavelength is given by the Rydberg formula:

where:

• $\lambda$ is the wavelength of emitted light,
• $R_H \approx 1.097 \times 10^7 \, \text{m}^{-1}$ is the Rydberg constant,
• $n$ is the lower orbit (the one we need),
• $m > n$ is the higher orbit.

The given wavelength $9546\,\text{Å}$ is in the infrared region (since visible light ends around $7000\,\text{Å}$). In the hydrogen spectrum, infrared lines belong to the Paschen series (transitions to $n=3$) or higher series. This immediately suggests that $n$ is likely 3, but we must verify.

Step-by-Step Solution

1. Convert wavelength to meters

2. Write the Rydberg formula

with $R_H = 1.097 \times 10^7 \, \text{m}^{-1}$.

3. Compute $1/\lambda$

4. Divide by $R_H$

So:

5. Test possible values of $n$

   • If $n=1$ (Lyman series): $1/n^2 = 1$. Then $1/m^2 = 1 - 0.0955 = 0.9045$ → $m \approx 1.05$, impossible since $m > n$.
   • If $n=2$ (Balmer series): $1/n^2 = 0.25$. Then $1/m^2 = 0.25 - 0.0955 = 0.1545$ → $m \approx 2.54$, not an integer.
   • If $n=3$ (Paschen series): $1/n^2 = 1/9 \approx 0.1111$. Then $1/m^2 = 0.1111 - 0.0955 = 0.0156$ → $m^2 \approx 64.1$ → $m \approx 8.01$, which is very close to 8.
   • If $n=4$ (Brackett series): $1/n^2 = 1/16 = 0.0625$. Then $1/m^2 = 0.0625 - 0.0955$ is negative, impossible.

   Only $n=3$ gives a plausible integer $m$ (specifically $m=8$).

6. Verify with exact calculation

   For $n=3$, $m=8$:

This matches perfectly.

The Lyman series corresponds to transitions in the hydrogen atom where an electron falls from a higher energy level ($n \geq 2$) to the ground state ($n = 1$). The wavelength of the emitted photon is given by the Rydberg formula:

where $R$ is the Rydberg constant. The first line of the Lyman series is the transition from $n=2$ to $n=1$; the third line is from $n=4$ to $n=1$. The ratio of wavelengths is simply the inverse ratio of the wavenumbers, so we compare $\frac{1}{\lambda_1}$ and $\frac{1}{\lambda_3}$.

1. First line ($n=2 \to 1$):

So $\lambda_1 = \frac{4}{3R}$.

2. Third line ($n=4 \to 1$):

So $\lambda_3 = \frac{16}{15R}$.

3. Ratio $\lambda_1 : \lambda_3$:

Thus $\lambda_1 : \lambda_3 = 5 : 4$.

Concept and Intuition

Classical electromagnetic theory states that any accelerating charged particle will radiate electromagnetic waves. An electron revolving in a circular orbit around a nucleus is constantly undergoing centripetal acceleration, even if its speed is constant. Therefore, according to classical theory, this orbiting electron should continuously emit electromagnetic radiation.

The frequency of the light emitted by such an accelerating charge is predicted to be equal to the frequency of its revolution. This is because the electron's position and velocity components vary periodically with the orbital frequency, and this periodic variation in charge distribution and current generates electromagnetic waves of the same frequency.

Step-by-step Derivation

1. Identify the goal: We need to find the frequency of light emitted by the electron according to classical electromagnetic theory. As discussed, this is equal to the orbital frequency of the electron.

2. Relate orbital velocity, radius, and frequency: For an object moving in a circular path, its linear velocity ($v$) is related to its angular velocity ($\omega$) and the radius of the orbit ($r$) by the formula $v = \omega r$. The angular velocity is, in turn, related to the orbital frequency ($f$) by $\omega = 2\pi f$.

3. Derive the formula for frequency: Combining these relationships, we get:

Rearranging this equation to solve for the frequency $f$:

> [!FORMULA]
> The orbital frequency $f$ of an electron moving with velocity $v$ in a circular orbit of radius $r$ is given by:
> $$f = \frac{v}{2\pi r}$$

4. Substitute the given values:

The radius of the second orbit of the hydrogen atom is given as $r = 2.12 \text{ Å}$. We need to convert this to meters:

$r = 2.12 \times 10^{-10} \text{ m}$

The velocity of the electron is given as $v = 1.1 \times 10^6 \text{ m/s}$.

5. Calculate the frequency: Substitute these values into the formula for $f$:

Using $\pi \approx 3.14159$:

6. Compare with the given options:

   (A) $2.05 \times 10^{14}$ Hz

   (B) $16.50 \times 10^{14}$ Hz

   (C) $4.15 \times 10^{14}$ Hz

   (D) $8.25 \times 10^{14}$ Hz

   Our calculated value $8.258 \times 10^{14}$ Hz is very close to option (D).

The key idea is that the wavelength of emitted radiation when an electron transitions between orbits in hydrogen is given by the Rydberg formula:

where $n_i$ is the initial orbit and $n_f$ is the final orbit. Since we only need the ratio of two wavelengths, the Rydberg constant $R$ cancels out, and we just compare the differences of inverse squares.

1. Write the formula for each transition For the transition from $n=4$ to $n=2$:

For the transition from $n=3$ to $n=2$:

2. Take the ratio of the wavelengths Since $\lambda$ is inversely proportional to the right-hand side, we have:

So the ratio is $20 : 27$.

3. Interpret the result The ratio $\lambda_{4\to2} : \lambda_{3\to2} = 20 : 27$ means the wavelength from the 4→2 transition is shorter (since 20 < 27), which makes sense because the energy difference is larger for 4→2 than for 3→2, so the wavelength is smaller.

The Bohr quantisation rule is $L=\dfrac{nh}{2\pi}$. Matching the given $\dfrac{3h}{2\pi}$ gives $n=3$, so the electron starts in the third orbit.

The emitted radiation comes from the downward transitions $3\to2$ and $3\to1$. Minimum wavelength means maximum photon energy, i.e. the largest value of $\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)$, which is the jump straight to the ground state ($n_i=3\to n_f=1$):

Concept & Intuition

In the hydrogen spectrum, each series is defined by a fixed lower energy level: Lyman series ends at n=1, Balmer at n=2. The longest wavelength in a series comes from the smallest possible energy jump — that is, from the nearest higher level to the series limit. For Lyman, that’s from n=2 to n=1; for Balmer, from n=3 to n=2. The Rydberg formula gives the inverse wavelength, so we can directly compare these two transitions.

Step-by-step solution

1. Recall the Rydberg formula for hydrogen:

where $R_H$ is the Rydberg constant, $n_f$ is the final level, and $n_i$ is the initial level ($n_i > n_f$).

2. Longest wavelength in the Lyman series ($n_f = 1$): The smallest energy transition is from $n_i = 2$ to $n_f = 1$.

So $\lambda_L = \frac{4}{3 R_H}$.

3. Longest wavelength in the Balmer series ($n_f = 2$): The smallest energy transition is from $n_i = 3$ to $n_f = 2$.

So $\lambda_B = \frac{36}{5 R_H}$.

4. Find the ratio $\lambda_L / \lambda_B$:

The key idea is the Bohr model of the hydrogen atom, which quantizes electron orbits. In this model, the radius of the $n$-th orbit is proportional to $n^2$, where $n$ is the principal quantum number. The ground state ($n=1$) has the smallest radius, called the Bohr radius $a_0$. The first excited state corresponds to $n=2$, so its radius is simply $4$ times the Bohr radius. No complex calculations are needed — just recall the scaling and the standard value of $a_0$.

1. Identify the state: The "first excited state" means the electron is not in the ground state but in the next energy level. In hydrogen, the ground state is $n=1$, so the first excited state is $n=2$.

2. Recall the Bohr radius formula: In the Bohr model, the radius of the $n$-th orbit is

where $a_0 = 0.529\,\text{Å}$ (the Bohr radius). This comes from balancing Coulomb force and centripetal force, plus quantization of angular momentum.

3. Apply for $n=2$:

Rounding to two decimal places gives $2.12\,\text{Å}$.

4. Match with options: Among the choices, $2.12\,\text{Å}$ corresponds to option (B).

The key idea here is that in the Bohr model of the hydrogen atom, the electron’s orbital period is not constant — it depends strongly on the principal quantum number $n$. The problem gives a ratio of periods, so we need to find how $T$ scales with $n$. Once we know that, the relation between $n_1$ and $n_2$ follows directly.

Why this approach works:

In the Bohr model, the electron moves in circular orbits with quantized angular momentum. The time period $T$ is the circumference divided by the orbital speed. Both the radius and the speed depend on $n$, so combining them gives a clean power law. This avoids having to compute actual numbers — just the scaling is enough.

Let’s work it out step by step.

1. Recall the Bohr model formulas for radius and speed. For a hydrogen-like atom (here, just hydrogen), the radius of the $n$-th orbit is

where $a_0$ is the Bohr radius. The orbital speed of the electron in that orbit is

where $v_0$ is the speed in the ground state ($n=1$). Both results come from balancing Coulomb force and centripetal force, plus the quantization condition $m v r = n \hbar$.

2. Express the time period in terms of $n$. The time period for one complete revolution is

Substitute the expressions:

The factor $\frac{2\pi a_0}{v_0}$ is a constant (the period of the ground state). So we have the scaling law:

3. Apply the given condition. The problem states: “The time period of the electron in the initial state is eight times that of the final state.” That is,

Using the proportionality $T \propto n^3$, this becomes

4. Solve for the ratio. Taking the cube root of both sides:

So the initial principal quantum number is twice the final one.