Problem 63 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Estimate the difference in energy between Ist and 2nd Bohr orbit for a hydrogen atom. At what minimum atomic number, a transition from \(n=2\) to \(n=1\) energy level would result in the emission of \(X\)-rays with \(\lambda=3.0 \times 10^{-8} \mathrm{~m} ?\) Which hydrogen atom-like species does this atomic number correspond to ?

Step-by-Step Solution

Answer

1) The energy difference between Ist and 2nd Bohr orbits is 10.2 eV. 2) The atomic number for X-ray emission is 2, corresponding to He⁺ ion.

Step 1: Understanding the Energy Formula for Hydrogen Atom

The energy of an electron in the nth orbit of a hydrogen atom is given by the formula \( E_n = -13.6 \frac{Z^2}{n^2} \) eV, where \( Z \) is the atomic number and \( n \) is the principal quantum number. For hydrogen, \( Z = 1 \).

Step 2: Calculating Energy for the 1st and 2nd Orbits

Using the energy formula, compute the energy of the electron in the 1st orbit (n=1) as \( E_1 = -13.6 \times 1^2 = -13.6 \) eV. For the 2nd orbit (n=2), the energy is \( E_2 = -13.6 \times \frac{1^2}{2^2} = -3.4 \) eV.

Step 3: Finding the Energy Difference Between Orbits

Calculate the energy difference \( \Delta E \) between the 1st and 2nd orbits using the equation \( \Delta E = E_2 - E_1 = (-3.4) - (-13.6) = 10.2 \) eV. This is the energy difference between the Ist and 2nd Bohr orbits for a hydrogen atom.

Step 4: Finding Energy for X-ray Emission

The energy of photons can also be calculated using \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \) Js, \( c = 3 \times 10^8 \) m/s, and \( \lambda = 3.0 \times 10^{-8} \) m. Solve to find \( E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.0 \times 10^{-8}} \approx 4.136 \times 10^{-15} \) Joules. Convert to eV: \( E \approx 4.136 \times 10^{-15} /1.602 \times 10^{-19} \approx 25.8 \) eV.

Step 5: Finding Minimum Atomic Number

Use the Rydberg formula \( E_{photon} = 13.6 \post{Z^2} \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \) for \( n_1=1 \), \( n_2=2 \). Setting \( E_{photon} = 25.8 \) eV leads to solving \( 25.8 = 13.6 \times Z^2\left(1 - \frac{1}{4}\right) \). Solve for \( Z \): \( Z^2 = \frac{25.8}{10.2} \approx 2.53 \). Thus, \( Z \approx 1.6 \). The nearest whole number is \( Z = 2 \).

Step 6: Identifying the Hydrogen-like Species

The atomic number \( Z = 2 \) corresponds to helium (He), which means this species could be a singly ionized helium ion, often referred to as \( He^+ \).

Key Concepts

Energy Levels

In the Bohr model, energy levels refer to the quantized orbits in which electrons reside around the nucleus of an atom. These energy levels are indicated by the principal quantum number, \( n \), which can take positive integer values like 1, 2, 3, and so forth. The energy associated with a specific level is given by the formula \( E_n = -13.6 \frac{Z^2}{n^2} \) eV, where \( Z \) represents the atomic number, typically 1 for hydrogen, and \( n \) signifies the orbit number.

• For \( n = 1 \), the electron is in the closest and most tightly bound orbit to the nucleus, resulting in larger negative energy.
• For \( n = 2 \), the electron is further from the nucleus which means its energy is less negative (or higher).

The difference in energy between these levels, calculated by subtracting the energies of one level from another, corresponds to a transition or jump that an electron makes between these levels. It is this quantization of the energy levels in Bohr's model that helps predict and explain phenomena such as spectral lines.

Atomic Number

The atomic number, denoted as \( Z \), is a fundamental property of an element and is equal to the number of protons in its nucleus. It uniquely identifies each chemical element and determines the element's place in the periodic table. In the Bohr model of an atom, the atomic number plays a crucial role because it influences the attractive force between the nucleus and the electrons, ultimately affecting the atom's energy levels.

• For hydrogen, \( Z = 1 \), meaning it has one proton in its nucleus.
• The energy levels depend on \( Z \) squared, as seen in the energy formula: \( E_n = -13.6 \frac{Z^2}{n^2} \) eV.

Higher atomic numbers lead to more tightly bound electrons, increasing the energy changes involved when electrons transition between levels. This principle becomes crucial in understanding the production of X-rays, where high-energy transitions require higher atomic numbers.

X-ray Emission

X-ray emission occurs when electrons transition between high energy levels, releasing electromagnetic radiation in the X-ray portion of the spectrum. When an electron falls from higher energy levels to closer orbits (such as from \( n = 2 \) to \( n = 1 \)), it emits energy in the form of X-ray photons, which have very short wavelengths and high energies.The Bohr model allows for the calculation of the energy of these X-rays utilizing the equation for photon energy: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted X-ray. X-ray emission is significant in various applications, including medical imaging and crystallography.For an atom to emit X-rays with a specific wavelength, its atomic number must be such that the energy released during electronic transitions matches the energy calculated from the given wavelength. This is how the minimum atomic number for X-ray emission is determined.

Hydrogen-like Species

Hydrogen-like species refer to ions of elements that have only one electron, similar to a hydrogen atom. These species can be derived from elements such as helium or lithium by removing all but one electron. They maintain a similar quantum mechanical simplicity to hydrogen, making them excellent candidates for study in quantum physics and spectroscopy.

• The simplest example of a hydrogen-like species is the \( He^+ \) ion, which is helium with one electron removed (thus leaving one).
• Their spectral lines and energy transitions can be directly calculated using formulas similar to those used for hydrogen since they each have a single electron.

Importantly, the energy levels of these species are dependent on the atomic number \( Z \), following \( E_n = -13.6 \frac{Z^2}{n^2} \) eV. This relationship implies that their spectral lines will shift with increasing atomic number, a phenomenon that helps expand their applications in various fields of science.