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

Question

Energy of an electron is given by, \(\mathrm{E}=-2.178 \times 10^{-18}\left(\frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}\right)\) 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 \(\left.\mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\) (a) \(6.500 \times 10^{-7} \mathrm{~m}\) (b) \(8.500 \times 10^{-7} \mathrm{~m}\) (c) \(1.214 \times 10^{-7} \mathrm{~m}\) (d) \(2.816 \times 10^{-7} \mathrm{~m}\)

Step-by-Step Solution

Answer

The wavelength required is approximately \(1.214 \times 10^{-7} \text{ m}\), option (c).

Step 1: Calculate Energy at n = 1

Using the formula for the energy of an electron, substitute the values: \( Z = 1 \) (for hydrogen) and \( n = 1 \). The energy \( E_1 \) is given by:\[E_1 = -2.178 \times 10^{-18} \left(\frac{1^2}{1^2}\right) = -2.178 \times 10^{-18} \, \text{Joules}\]

Step 2: Calculate Energy at n = 2

Similarly, calculate the energy of an electron in the hydrogen atom at \( n = 2 \):\[E_2 = -2.178 \times 10^{-18} \left(\frac{1^2}{2^2}\right)\]\[E_2 = -2.178 \times 10^{-18} \times \frac{1}{4} = -0.5445 \times 10^{-18} \, \text{Joules}\]

Step 3: Determine Energy Difference

The energy required to excite the electron from \( n = 1 \) to \( n = 2 \) is the difference between \( E_2 \) and \( E_1 \):\[\Delta E = E_2 - E_1 = (-0.5445 \times 10^{-18}) - (-2.178 \times 10^{-18})\]\[\Delta E = 1.6335 \times 10^{-18} \, \text{Joules}\]

Step 4: Find Wavelength using the Energy-Wavelength Relation

Use the relation between energy and wavelength: \( \Delta E = \frac{hc}{\lambda} \). Solve for \( \lambda \):\[\lambda = \frac{hc}{\Delta E}\]Substitute \( h = 6.62 \times 10^{-34} \, \text{Js} \), \( c = 3.0 \times 10^{8} \, \text{m/s} \), and \( \Delta E = 1.6335 \times 10^{-18} \, \text{Joules} \):\[\lambda = \frac{6.62 \times 10^{-34} \times 3.0 \times 10^{8}}{1.6335 \times 10^{-18}}\]\[\lambda = 1.215 \times 10^{-7} \, \text{m}\]

Step 5: Select the Closest Option

The calculated wavelength is \( 1.215 \times 10^{-7} \text{ m} \), which is closest to option (c) \( 1.214 \times 10^{-7} \text{ m} \).

Key Concepts

Electron Excitation

Electron excitation occurs when an electron in an atom absorbs energy and moves from a lower energy level to a higher one. In the hydrogen atom, this can be visualized by an electron jumping from one shell to another.
The energy levels are designated by the quantum number 'n'. The smallest value of n is 1, representing the ground state, where the electron is nearest to the nucleus and has the least energy.

• When it absorbs energy, it can move to a higher level (n>1), such as n=2, n=3, etc. This process is called excitation.
• The energy difference between the two levels determines the amount of energy absorbed.

In our example, moving from level n=1 to n=2 involves energy absorption to excite the electron from the lower ( 1) to the higher ( 2) state.

Energy Calculation

To determine the necessary energy for electron excitation in a hydrogen atom, we need to use the formula for electron energy: \[E = -2.178 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right)\] where \(E\) is the energy, \(Z\) is the atomic number (1 for hydrogen), and \(n\) is the principal quantum number.
First, calculate the energy at the initial level \(n = 1\):\[E_1 = -2.178 \times 10^{-18} \times \frac{1^2}{1^2} = -2.178 \times 10^{-18} \, ext{Joules}\]
Then, the energy at the final level \(n = 2\): \[E_2 = -2.178 \times 10^{-18} \times \frac{1^2}{2^2} = -0.5445 \times 10^{-18} \, ext{Joules}\]

• Subtract \(E_1\) from \(E_2\) to find the energy difference \(\Delta E\), required for excitation: \[\Delta E = E_2 - E_1 = 1.6335 \times 10^{-18} \, \text{Joules}\]

This energy difference represents the precise amount of energy needed to move the electron from n=1 to n=2.

Wavelength of Light

The wavelength of light associated with the energy transition of an electron can be calculated using the energy-wavelength relationship: \[\Delta E = \frac{hc}{\lambda}\]where \(h\) is Planck's constant \(6.62 \times 10^{-34} \, \text{Js}\), \(c\) is the speed of light \(3.0 \times 10^8 \, \text{m/s}\), and \(\lambda\) is the wavelength.
Solve for \(\lambda\):\[\lambda = \frac{hc}{\Delta E}\]

• Insert the known values: \(\lambda = \frac{6.62 \times 10^{-34} \times 3.0 \times 10^8}{1.6335 \times 10^{-18}}\)

This gives the result: \[\lambda = 1.215 \times 10^{-7} \, ext{m}\]
The calculated wavelength matches the closest choice given in the original exercise, confirming this method effectively relates energy changes in atomic transitions to observable light wavelengths.