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By what percentage is the electron mass changed in using the reduced mass for the hydrogen atom? What would the reduced mass be if the proton had the same mass as the electron?

Short Answer

Expert verified
Answer: The percentage change in the electron mass when using the reduced mass for the hydrogen atom is 0.0502%, and the reduced mass would be 4.55469178 × 10^-31 kg if the proton had the same mass as the electron.

Step by step solution

01

Find the reduced mass of the hydrogen atom

To do this, we'll use the formula for the reduced mass of a two-body system, given by: reduced mass (µ) = (m1 * m2) / (m1 + m2) where m1 and m2 are the masses of the two interacting particles (proton and electron in this case). The values we have: mass of electron (me) = 9.10938356 × 10^-31 kg mass of proton (mp) = 1.67262192 × 10^-27 kg Plug the values into the formula, µ = (me * mp) / (me + mp)
02

Calculate the percentage change in the electron mass

Now that we've found the reduced mass, we can calculate the difference between the electron mass and the reduced mass and then find the percentage change. percentage change = ((me - µ) / me) * 100
03

Find the reduced mass if the proton had the same mass as the electron

In this part, we'll assume that the proton has the same mass as the electron (me). So, the formula for the reduced mass will be: µ = (me * me) / (me + me) Now, let's calculate the values for each step.
04

calculations

µ = (9.10938356 × 10^-31 kg * 1.67262192 × 10^-27 kg) / (9.10938356 × 10^-31 kg + 1.67262192 × 10^-27 kg) = 9.10481522 × 10^-31 kg
05

calculations

percentage change = ((9.10938356 × 10^-31 kg - 9.10481522 × 10^-31 kg) / 9.10938356 × 10^-31 kg) * 100 = 0.0502 %
06

calculations

µ = (9.10938356 × 10^-31 kg * 9.10938356 × 10^-31 kg) / (9.10938356 × 10^-31 kg + 9.10938356 × 10^-31 kg) = 4.55469178 × 10^-31 kg So, the electron mass changes by 0.0502 % when using the reduced mass for the hydrogen atom, and the reduced mass would be 4.55469178 × 10^-31 kg if the proton had the same mass as the electron.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Hydrogen Atom
Imagine a tiny system where two particles dance around each other in the vast emptiness of space - that's how we can visualize the hydrogen atom. At its simplest, a hydrogen atom consists of just one electron and one proton. The proton sits at the core, while the electron orbits around it.

Although the proton is significantly heavier than the electron, its mass is a key factor in the behavior of the system. In quantum physics, the hydrogen atom serves as a fundamental model for understanding larger, more complex atoms and their interactions. It's the simplicity of this single-electron atom that allows scientists to make precise calculations about its properties, making the hydrogen atom a cornerstone in the field of atomic physics.
Electron Mass Significance
The mass of an electron is incredibly small, at approximately 9.11 x 10^-31 kilograms. Despite its minuscule weight, the electron's mass has a profound impact on the nature of chemical bonds and atomic structures. For example, the mass determines the size of an atom and the energy levels within an atom that electrons can occupy.

In our explorations of atomic and subatomic physics, the accurate value of the electron mass is essential. It also plays a pivotal role in defining the strength of the bond between the electron and the nucleus in various materials, which has direct implications on the material's properties.
Proton Mass Relevance
When we weigh a proton, we find it has a mass of about 1.67 x 10^-27 kilograms, nearly 1836 times the mass of an electron. This mightiness is what anchors the electrons in an atom, leading to a relatively stable system.

The proton mass is much more than just a number; it affects the balance of forces within the nucleus and thus defines how atoms come together to form molecules. The mass of protons also influences the energy levels within the nucleus and plays a critical role in nuclear reactions. Any change in the proton mass would drastically alter the chemistry we observe in the universe.
The Two-Body System and Reduced Mass
The concept of a two-body system arises when two particles, such as an electron and a proton, influence each other through a force, like the electromagnetic force in an atom. Here, the notion of reduced mass comes into play as a simplifying tool.

The reduced mass, denoted by Greek letter mu (µ), is a way to simplify calculations when two bodies orbit around a common center of mass. It is calculated as the product of the two masses divided by their sum. The beauty of the reduced mass concept is that it allows physicists to treat a two-body problem as if it were a single-body problem with this adjusted mass. In atomic systems, this simplification is invaluable for solving complex equations describing the motion of electrons around nuclei.
Percentage Change Calculation
Percentage change is a mathematical concept used to describe the extent of change over time or between different conditions, expressed as a percentage. It's calculated by taking the difference between the old and new value, divided by the old value, and then multiplying by 100 to get a percentage.

In the context of our exercise, we used the percentage change calculation to understand the impact of using the reduced mass on the effective mass of the electron in a hydrogen atom. By showing a percentage change of just 0.0502%, we reveal that the difference is marginal but nonetheless significant in precise scientific calculations, where even small variations can lead to different outcomes in experimental data and theoretical models.

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Most popular questions from this chapter

Calculate the energy needed to change a single ionized helium atom into a double ionized helium atom (that is, change it from \(\mathrm{He}^{+}\) into \(\mathrm{He}^{2+}\) ). Compare it to the energy needed to ionize the hydrogen atom. Assume that both atoms are in their fundamental state.

An electron in a hydrogen atom is in the ground state (1s). Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground state wave function for hydrogen is: \(\psi_{1 s}(r)=A_{1 s} e^{-r / a_{0}}=e^{-r / a_{0}} / \sqrt{\pi a_{0}^{3}}\).

The radius of the \(n=1\) orbit in the hydrogen atom is $$ a_{0}=0.053 \mathrm{nm} $$ a) Compute the radius of the \(n=6\) orbit. How many times larger is this compared to the \(n=1\) radius? b) If an \(n=6\) electron relaxes to an \(n=1\) orbit (ground state), what is the frequency and wavelength of the emitted radiation? What kind of radiation was emitted (visible, infrared, etc.)? c) How would your answer in (a) change if the atom was a singly ionized helium atom \(\left(\mathrm{He}^{+}\right),\) instead?

An electron in a hydrogen atom is in the \(2 s\) state. Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground-state wave function for hydrogen is: $$ \psi_{2 s}(r)=\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}}. $$ The integral is a bit tedious, so you may want consider using mathematical programs such as Mathcad, Mathematica, etc., or doing the integral online at http://integrals.wolfram.com/index.jsp.

An 8.00 -eV photon is absorbed by an electron in the \(n=2\) state of a hydrogen atom. Calculate the final speed of the electron.

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