Physics - Modern Physics Question with Solution | TestHub

The wavelength of the first Lyman series will be maximum for the hydrogen atom.

Explanation:

The wavelength of emitted radiation in a hydrogen-like atom during electronic transitions is given by the Rydberg formula:

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

where:

- $\lambda$ is the wavelength of the emitted photon

- $R$ is the Rydberg constant ($R\approx 1.097\times 10^7{m}^{-1}$)

- $Z$ is the atomic number (number of protons in the nucleus)

- $n_1$ and $n_2$ are the principal quantum numbers of the initial and final energy levels, respectively ($n_2>n_1$)

For the Lyman series, the electron transitions occur from $n_2>1$ to $n_1=1$. The first Lyman series corresponds to the transition from $n_2=2$ to $n_1=1$. Therefore,

$$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R{Z}^2\left(1-\frac{1}{4}\right)=\frac{3}{4}R{Z}^2$$

$$\lambda =\frac{4}{3R{Z}^2}$$

To maximize the wavelength $\lambda$, we need to minimize $Z^2$, which means minimizing $Z$.

A. Hydrogen atom (H): $Z=1$

B. Deuterium atom (D): $Z=1$ (Deuterium is an isotope of hydrogen, so it has the same atomic number)

C. Helium ion (He⁺): $Z=2$

D. Lithium ion (Li²⁺): $Z=3$

Since both hydrogen and deuterium have the smallest atomic number ($Z=1$), they will have the maximum wavelength. However, deuterium has a slightly heavier nucleus than hydrogen. The Rydberg constant $R$ is slightly affected by the mass of the nucleus. A heavier nucleus results in a slightly larger Rydberg constant, and thus a slightly smaller wavelength. Since the question does not account for this small difference, both A and B are technically correct, but A is the standard answer.

Comparing the given options:

- Hydrogen (Z=1): $\lambda =\frac{4}{3R(1{)}^2}=\frac{4}{3R}$

- Deuterium (Z=1): $\lambda \approx \frac{4}{3R^{\prime }}$ where $R^{\prime }>R$ (slightly smaller wavelength than hydrogen)

- Helium ion (Z=2): $\lambda =\frac{4}{3R(2{)}^2}=\frac{4}{12R}=\frac{1}{3R}$

- Lithium ion (Z=3): $\lambda =\frac{4}{3R(3{)}^2}=\frac{4}{27R}$

Therefore, the wavelength is maximum for the hydrogen atom.

Final Answer: The final answer is $\boxed{A}$