Problem 178 - A Complete Resource Book in Chemistry for JEE Main

Question

Energy of an electron is given by, [2013] \(\mathrm{E}=-2.178 \times 10^{-18}\left(\frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}\right)\) Wavelength of light required to excite an electron in an hydrogen atom from level \(\mathrm{n}=1\) to \(\mathrm{n}=2\) will be \(\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}\right.\) and \(\left.\mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\)Energy of an electron is given by, [2013] \(\mathrm{E}=-2.178 \times 10^{-18}\left(\frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}\right)\) Wavelength of light required to excite an electron in an hydrogen atom from level \(\mathrm{n}=1\) to \(\mathrm{n}=2\) will be \(\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}\right.\) and \(\left.\mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\)

Step-by-Step Solution

Answer

The wavelength is 121.5 nm.

Step 1: Calculate Initial Energy (n=1)

First, calculate the energy when the electron is in level \(n=1\). Use the formula \(E = -2.178 \times 10^{-18} \times \frac{Z^2}{n^2}\). Here, \(Z=1\) for hydrogen and \(n=1\). This gives: \[E_1 = -2.178 \times 10^{-18} \times \frac{1^2}{1^2} = -2.178 \times 10^{-18} \, J\].

Step 2: Calculate Final Energy (n=2)

Next, calculate the energy when the electron is in level \(n=2\). Again, use the formula \(E = -2.178 \times 10^{-18} \times \frac{Z^2}{n^2}\). For \(n=2\): \[E_2 = -2.178 \times 10^{-18} \times \frac{1^2}{2^2} = -2.178 \times 10^{-18} \times \frac{1}{4} = -0.5445 \times 10^{-18} \, J\].

Step 3: Calculate Energy Difference

Find the energy difference \(\Delta E\) required to move the electron from \(n=1\) to \(n=2\). This is \(E_2 - E_1\): \[\Delta E = E_2 - E_1 = (-0.5445 \times 10^{-18}) - (-2.178 \times 10^{-18}) = 1.6335 \times 10^{-18} \, J\].

Step 4: Calculate Wavelength using Energy-Wavelength Relation

Use the formula \(\Delta E = \frac{hc}{\lambda}\) to find the wavelength \(\lambda\). Rearrange the formula to \(\lambda = \frac{hc}{\Delta E}\). Substitute \(h = 6.62 \times 10^{-34} \, Js\), \(c = 3.0 \times 10^8 \, m/s\), and \(\Delta E = 1.6335 \times 10^{-18} \, J\): \[ \lambda = \frac{(6.62 \times 10^{-34})(3.0 \times 10^8)}{1.6335 \times 10^{-18}} \]. This calculates to \(\lambda = 1.215 \times 10^{-7} \, m\) or \(121.5 \, nm\).

Key Concepts

Bohr model

To understand hydrogen atom energy levels, one must first delve into the Bohr model. Proposed by Niels Bohr in 1913, this model revolutionized atomic physics by introducing quantized energy levels for electrons orbiting the nucleus. Bohr suggested that electrons orbit the nucleus at specific distances, forming distinct levels. Each level corresponds to a specific energy state.
This model is based on the idea that electrons can only occupy certain orbits where their angular momentum is quantized. Such quantization implies that electrons do not radiate energy, allowing them to remain stable in these fixed paths.

• The closer the electron is to the nucleus, the lower its energy position.
• Energy levels are quantized, meaning electrons can only exist in particular levels, not in between.
• This model primarily applies to hydrogen and hydrogen-like atoms, where there is only a single electron.

Despite its limitations with multi-electron atoms, the Bohr model successfully explained the discrete wavelengths of spectral lines.

Electron transition

In the Bohr model, transitions between energy levels occur when an electron absorbs or releases energy. This phenomenon, known as an electron transition, results in the emission or absorption of a photon with a specific wavelength.
A transition only happens when an electron moves from one quantized energy level to another. The key points to note here are:

• If an electron jumps from a lower to a higher energy level (excitation), it absorbs energy.
• When an electron moves from a higher to a lower energy level (relaxation), it emits energy.
• The energy of the absorbed or emitted photon matches the energy difference between the two levels.

In our exercise, the transition from level =1 to =2 represents excitation. This requires an input of energy, typically from absorbing a photon that exactly matches the energy gap between the levels.

Energy calculation

To determine the energy required for an electron transition, we can use a specific equation derived from the Bohr model. It is important to remember:

• The energy of an electron in a given level is calculated using: \(E = -2.178 \times 10^{-18} \times \frac{Z^2}{n^2}\) where \(Z\) is the atomic number (for hydrogen, \(Z=1\)).
• The energy is always negative, implying that electrons are bound to the nucleus.
• The energy needed for a transition is the difference between the initial and final energy levels.

In the exercise, you see two calculations. First for \(n=1\), yielding \(E_1 = -2.178 \times 10^{-18} \, J\). Second for \(n=2\), yielding \(E_2 = -0.5445 \times 10^{-18} \, J\). The energy difference or \(\Delta E\) is \(1.6335 \times 10^{-18} \, J\), which matches the energy of a photon needed for this transition.

Wavelength calculation

Once the energy difference for an electron transition is known, the wavelength of the light associated with the transition can be calculated. The relationship between energy and wavelength for photons is given by:

• The equation \(\Delta E = \frac{hc}{\lambda}\) links energy difference (\(\Delta E\)) to wavelength (\(\lambda\)).
• \(h\) is Planck's constant \((6.62 \times 10^{-34} \, Js)\), and \(c\) is the speed of light \((3.0 \times 10^8 \, m/s)\).
• This formula shows that the energy of a photon is inversely proportional to its wavelength.

Rearranging the formula provides the required wavelength as \(\lambda = \frac{hc}{\Delta E}\). For the given energy difference \(1.6335 \times 10^{-18} \, J\), the calculated wavelength is \(1.215 \times 10^{-7} \, m\), which equates to \(121.5 \, nm\). This calculation shows precisely the wavelength of the photon absorbed during the electron's transition from \(n=1\) to \(n=2\).