Let's delve into the questions regarding the mixture of hydrogen (H) and singly ionized helium (He⁺) ions, focusing on their energy states and the resulting emissions. We will analyze each question step by step, utilizing the Bohr model of the atom, which simplifies our calculations for hydrogen-like atoms.

Understanding the Quantum States of He⁺

In the scenario presented, we have H atoms and He⁺ ions that are excited to their first excited states. The Bohr model provides a framework for understanding the energy levels of these hydrogen-like atoms. For He⁺, which has only one electron, the energy levels can be described by the formula:

E_n = -\frac{Z^2 \cdot R_H}{n^2}

Here, Z is the atomic number (which is 2 for He), R_H is the Rydberg constant (approximately 13.6 eV for hydrogen), and n is the principal quantum number.

Question 41: Quantum Number of the State in He⁺

When H atoms transfer their excitation energy to He⁺ ions, we need to determine the quantum number n of the state that is populated in He⁺. Since H atoms can reach their first excited state (n=2), they can transfer energy equivalent to the difference between the energy levels of He⁺.

For He⁺, the energy levels are:

• n = 1: E₁ = -\frac{2^2 \cdot 13.6}{1^2} = -54.4 eV
• n = 2: E₂ = -\frac{2^2 \cdot 13.6}{2^2} = -13.6 eV
• n = 3: E₃ = -\frac{2^2 \cdot 13.6}{3^2} \approx -6.02 eV
• n = 4: E₄ = -\frac{2^2 \cdot 13.6}{4^2} \approx -3.4 eV
• n = 5: E₅ = -\frac{2^2 \cdot 13.6}{5^2} \approx -2.18 eV

The energy transferred from H (which is approximately 10.2 eV for the transition from n=1 to n=2) can excite He⁺ to n=2. Therefore, the quantum number n of the state finally populated in He⁺ ions is 2.

Question 42: Wavelength of Light Emitted

Next, we need to calculate the wavelength of light emitted by He⁺ ions after they are excited to the n=2 state and then transition back to a lower energy state. The transition from n=2 to n=1 will emit a photon, and we can use the Rydberg formula to find the wavelength:

\(\frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)

Substituting the values for He⁺ (Z=2, n₁=1, n₂=2):

\(\frac{1}{\lambda} = 13.6 \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 54.4 \left( 1 - \frac{1}{4} \right) = 54.4 \cdot \frac{3}{4} = 40.8\)

Thus, the wavelength \(\lambda\) is:

\(\lambda = \frac{1}{40.8} \approx 2.45 \times 10^{-2} m\)

However, we need to convert this to nanometers for better clarity. The visible region typically ranges around 400 nm to 700 nm. The correct wavelength emitted in the visible region from the options provided is 5.6 × 10^-7 m.

Question 43: Kinetic Energy Ratio

Finally, we need to find the ratio of the kinetic energy of the n=2 electron for the H atom to that of the He⁺ ion. The kinetic energy (KE) of an electron in a hydrogen-like atom can be expressed as:

KE = \frac{Z^2 \cdot R_H}{2n^2}

For H (Z=1, n=2):

KE_H = \frac{1^2 \cdot 13.6}{2 \cdot 2^2} = \frac{13.6}{8} = 1.7 eV

For He⁺ (Z=2, n=2):

KE_He+ = \frac{2^2 \cdot 13.6}{2 \cdot 2^2} = \frac{54.4}{8} = 6.8 eV

Now, the ratio of the kinetic energies is:

Ratio = \frac{KE_H}{KE_He+} = \frac{1.7}{6.8} \approx 0.25

Thus, the ratio of the kinetic energy of the n=2 electron for the H atom to that of He⁺ is approximately 1:4.

In summary, the answers to the questions are:

• Q41: 2
• Q42: 5.6 × 10^-7