Problem 64
Question
The electron energy in hydrogen atom is given by \(E=\left(-21.7 \times 10^{-12}\right) / n^{2}\) ergs. Calculate the energy required to remove an electron completely from the \(n=2\) orbit. What is the longest wavelength (in \(\mathrm{cm}\) ) of light that can be used to cause this transition?
Step-by-Step Solution
Verified Answer
The energy required is \(5.425 \times 10^{-12}\) ergs, and the longest wavelength is \(3.66 \times 10^{-5}\) cm.
1Step 1: Identify Key Information
The formula for electron energy in a hydrogen atom is given as \(E=\left(-21.7 \times 10^{-12}\right) / n^{2}\) ergs. We need to calculate the energy when \(n=2\), and find the longest wavelength of light causing this transition.
2Step 2: Calculate Energy for n=2
Substitute \(n=2\) into the energy formula: \(E=\left(-21.7 \times 10^{-12}\right) / 2^{2} = \left(-21.7 \times 10^{-12}\right) / 4 = -5.425 \times 10^{-12}\) ergs.
3Step 3: Determine Energy Required to Remove the Electron
To completely remove the electron, its total energy must become zero. The energy required is the absolute value of the current energy level. So the required energy is \(5.425 \times 10^{-12}\) ergs.
4Step 4: Convert Energy to Wavelength
The energy and wavelength are related by the formula \(E = \frac{hc}{\lambda}\), where \(h = 6.626\times 10^{-27}\) erg·s is Planck's constant and \(c = 3\times10^{10}\) cm/s is the speed of light. Rearranging for \(\lambda\), we get \(\lambda = \frac{hc}{E}\).
5Step 5: Calculate Longest Wavelength
Substitute the values into the rearranged formula: \(\lambda = \frac{6.626\times 10^{-27}\ \text{erg·s} \times 3\times10^{10}\ \text{cm/s}}{5.425 \times 10^{-12}\ \text{erg}} = 3.66 \times 10^{-5}\) cm.
Key Concepts
Electron TransitionWavelength CalculationEnergy Quantization
Electron Transition
When we talk about electron transition in a hydrogen atom, we refer to the change in the energy state of an electron. Electrons orbit the nucleus in defined energy levels, denoted by quantum numbers (n). An electron in hydrogen can transition between these levels by absorbing or emitting energy.
- If an electron absorbs energy, it moves to a higher energy level or farther from the nucleus, like from n=2 to n=3.
- If an electron emits energy, it falls to a lower energy level, such as from n=3 to n=2.
Wavelength Calculation
Wavelength is a measure of the distance between consecutive peaks of a wave. In the context of electron transitions, each energy transition has a corresponding wavelength of light absorbed or emitted. To find this wavelength, we use the relation between energy and wavelength given by the formula: \[ E = \frac{hc}{\lambda} \]Here, \( E \) is the energy change, \( h \) is Planck's constant, and \( c \) is the speed of light. To find the wavelength \( \lambda \), we rearrange the equation to: \[ \lambda = \frac{hc}{E} \]By substituting the known values of \( h \), \( c \), and the calculated energy \( E \), we can determine the wavelength, which tells us the type of light required for the transition.
Energy Quantization
Energy quantization is a fundamental concept in quantum mechanics. For electrons in a hydrogen atom, energy levels are quantized, meaning they can only exist in specific discrete states. These states are defined by the principal quantum number, \( n \).
- The lower the energy level (smaller \( n \)), the more tightly bound the electron is to the nucleus.
- Higher energy levels (larger \( n \)) are less tightly bound, allowing electrons to be more excited or even removed from the atom.
Other exercises in this chapter
Problem 62
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition \(n=4\) to \(n=2\) of \(\mathrm{He}^{+}\)spectrum?
View solution Problem 63
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
View solution Problem 65
Naturally occurring boron consists of two isotopes whose atomic weights are \(10.01\) and \(11.01\). The atomic weight of natural boron is \(10.81\). Calculate
View solution Problem 61
Find out the number of waves made by a Bohr electron in one complete revolution in its 3rd orbit.
View solution