Question

calculate maximum wavelength of radiation emitted, producing a line in balmer series when electron fall from fourth energy level of He+ ion

Answer 1

121.53 nm

Answer 2

To calculate the maximum wavelength of radiation emitted, producing a line in the Balmer series when an electron falls from the fourth energy level of a He+ ion, we use the Rydberg formula for hydrogen-like atoms.

1. Identify the ion and its atomic number (Z):
   The ion is He+, which is a hydrogen-like atom. Its atomic number, $Z = 2$.

2. Identify the final energy level ($n_f$):
   A line in the Balmer series corresponds to electron transitions to the second energy level. So, $n_f = 2$.

3. Identify the initial energy level ($n_i$):
   The electron falls from the fourth energy level. So, $n_i = 4$.
   Note: The term "maximum wavelength" in this context refers to the wavelength of this specific transition, as the initial and final energy levels are explicitly given. If the question asked for the maximum wavelength in the Balmer series without specifying the initial level, then $n_i=3$ (the lowest possible initial level for Balmer series) would be used.

4. Apply the Rydberg formula:
   The formula for the wavelength of emitted radiation for hydrogen-like atoms is: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$
   Where:

• $\lambda$ is the wavelength of the emitted radiation.
• $R$ is the Rydberg constant ($1.097 \times 10^7 \text{ m}^{-1}$).
• $Z$ is the atomic number.
• $n_f$ is the final principal quantum number.
• $n_i$ is the initial principal quantum number.

5. Substitute the values:
   $\frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \times (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$
   $\frac{1}{\lambda} = (1.097 \times 10^7) \times 4 \left( \frac{1}{4} - \frac{1}{16} \right)$
   $\frac{1}{\lambda} = (1.097 \times 10^7) \times 4 \left( \frac{4}{16} - \frac{1}{16} \right)$
   $\frac{1}{\lambda} = (1.097 \times 10^7) \times 4 \left( \frac{3}{16} \right)$
   $\frac{1}{\lambda} = (1.097 \times 10^7) \times \frac{12}{16}$
   $\frac{1}{\lambda} = (1.097 \times 10^7) \times \frac{3}{4}$
   $\frac{1}{\lambda} = 1.097 \times 0.75 \times 10^7 \text{ m}^{-1}$
   $\frac{1}{\lambda} = 0.82275 \times 10^7 \text{ m}^{-1}$

6. Calculate the wavelength ($\lambda$):
   $\lambda = \frac{1}{0.82275 \times 10^7} \text{ m}$
   $\lambda = \frac{10^{-7}}{0.82275} \text{ m}$
   $\lambda \approx 1.2153 \times 10^{-7} \text{ m}$

7. Convert to nanometers (nm) for convenience:
   $1 \text{ m} = 10^9 \text{ nm}$
   $\lambda \approx 1.2153 \times 10^{-7} \times 10^9 \text{ nm}$
   $\lambda \approx 121.53 \text{ nm}$

The maximum wavelength of radiation emitted for the specified transition is approximately 121.53 nm.