परमाणु एवं नाभिक

According to Bohr's model of the hydrogen atom, for an electron orbiting in a stationary orbit:

• Kinetic energy: $K = \frac{1}{8\pi\varepsilon_0} \frac{e^2}{r} \gt 0$
• Potential energy: $U = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} = -2K$
• Total mechanical energy: $E = K + U = K + (-2K) = -K$
  Given that kinetic energy is $K$, the potential energy is $-2K$ and the total energy is $-K$.

The shortest wavelength (series limit, $n_2 = \infty$) for a spectral series with lower level $n_1$ is given by Rydberg's formula:

$$\frac{1}{\lambda_{\min}} = R_H \left(\frac{1}{n_1^2} - \frac{1}{\infty^2}\right) = \frac{R_H}{n_1^2} \implies \lambda_{\min} = \frac{n_1^2}{R_H}$$

Calculating the shortest wavelength for each series:

• Lyman series ($n_1 = 1$): $\lambda_{\min} = \frac{1}{R_H} = \frac{1}{R_H}$
• Balmer series ($n_1 = 2$): $\lambda_{\min} = \frac{4}{R_H}$
• Brackett series ($n_1 = 4$): $\lambda_{\min} = \frac{16}{R_H}$
• Pfund series ($n_1 = 5$): $\lambda_{\min} = \frac{25}{R_H}$
  Arranging in decreasing order of wavelength (largest to smallest):
  $$\text{Pfund (A)} \gt \text{Brackett (C)} \gt \text{Balmer (B)} \gt \text{Lyman (D)}$$
  $$(A), (C), (B), (D)$$

From the energy level diagram of hydrogen atom, the transition energy $\Delta E$ for each emission is:

• Transition A (from $-0.85\,\text{eV}$ to $-1.5\,\text{eV}$):

$$\Delta E_A = -0.85 - (-1.5) = 0.65\,\text{eV}$$

• Transition B (from $-0.85\,\text{eV}$ to $-3.4\,\text{eV}$):

$$\Delta E_B = -0.85 - (-3.4) = 2.55\,\text{eV}$$

• Transition C (from $-1.5\,\text{eV}$ to $-3.4\,\text{eV}$):

$$\Delta E_C = -1.5 - (-3.4) = 1.90\,\text{eV}$$

• Transition D (from $-1.5\,\text{eV}$ to $-13.6\,\text{eV}$):

$$\Delta E_D = -1.5 - (-13.6) = 12.10\,\text{eV}$$

Comparing transition energies:

$$\Delta E_D \gt \Delta E_B \gt \Delta E_C \gt \Delta E_A$$

Since wavelength is inversely proportional to transition energy ($\lambda = \frac{hc}{\Delta E}$), the order of wavelengths is:

$$\lambda_A \gt \lambda_C \gt \lambda_B \gt \lambda_D$$