This problem demonstrates that the binding energy of the electron in the ground state of a hydrogen atom is much smaller than the rest mass energies of the proton and electron. (a) Calculate the mass equivalent in u of the \(13.6-\mathrm{eV}\) binding energy of an electron in a hydrogen atom, and compare this with the known mass of the hydrogen atom. (b) Subtract the known mass of the proton from the known mass of the hydrogen atom. (c) Take the ratio of the binding energy of the electron (13.6 eV) to the energy equivalent of the electron's mass (0.511 \(\mathrm{MeV}\) ). (d) Discuss how your answers confirm the stated purpose of this problem.

Short Answer

Expert verified
The mass equivalent of the 13.6 eV binding energy is approximately 1.46 x 10^{-8} u, which is much smaller than the mass of the hydrogen atom (1.007825 u) and the mass of the proton (1.007276 u). The mass difference between the hydrogen atom and the proton is 0.000549 u, still much larger than the mass equivalent of the binding energy. Additionally, the ratio of the binding energy (13.6 eV) to the energy equivalent of the electron's mass (511,000 eV) is 2.66 x 10^{-5}, indicating that the binding energy is considerably smaller than the electron's rest mass energy. This confirms that the binding energy of the electron in the ground state of a hydrogen atom is much smaller than the rest mass energies of the proton and electron.

Step by step solution

01

Calculate the mass equivalent of the binding energy in atomic mass units (u)

First, we need to convert the given electron binding energy (13.6 eV) to atomic mass units (u). To do this, we will need the conversion factors: 1 u = 931.5 MeV 1 eV = 1 x 10^{-6} MeV The mass equivalent in u of the 13.6 eV binding energy can be calculated using these conversion factors: Mass equivalent (u) = binding energy (MeV) / 931.5 MeV/u Binding energy (MeV) = 13.6 eV * 1 x 10^{-6} MeV/eV ≈ 1.36 x 10^{-5} MeV Mass equivalent (u) ≈ 1.36 x 10^{-5} MeV / 931.5 MeV/u ≈ 1.46 x 10^{-8} u. Now, we can compare this calculated mass equivalent with the known mass of the hydrogen atom: Mass of hydrogen atom = 1.007825 u
02

Subtract the mass of the proton from the mass of the hydrogen atom

Next, we will subtract the known mass of the proton from the known mass of the hydrogen atom to find the difference: Mass difference = Mass of hydrogen atom - Mass of proton Mass of proton = 1.007276 u Mass difference ≈ 1.007825 u - 1.007276 u ≈ 0.000549 u
03

Calculate the ratio of the binding energy to the energy equivalent of the electron's mass

Now, we will find the ratio of the given electron binding energy (13.6 eV) to the energy equivalent of the electron's mass (0.511 MeV): Ratio = binding energy (eV) / energy equivalent of the electron's mass (eV) Energy equivalent of electron's mass = 0.511 MeV * 1 x 10^6 eV/MeV = 511000 eV Ratio ≈ 13.6 eV / 511000 eV ≈ 2.66 x 10^{-5}
04

Discuss the findings

The results obtained in the previous steps demonstrate that the binding energy of the electron in a hydrogen atom is much smaller than the rest mass energies of the proton and electron. The mass equivalent of the binding energy is approximately 1.46 x 10^{-8} u, which is significantly smaller than the mass of the hydrogen atom (1.007825 u) and the mass of the proton (1.007276 u). The mass difference between the hydrogen atom and the proton is 0.000549 u, which is still much larger than the mass equivalent of the binding energy. Furthermore, the ratio of the binding energy (13.6 eV) to the energy equivalent of the electron's mass (511,000 eV) is 2.66 x 10^{-5}, indicating that the binding energy is much smaller than the electron's rest mass energy. In conclusion, this exercise successfully demonstrates that the binding energy of the electron in the ground state of a hydrogen atom is much smaller when compared to the rest mass energies of the proton and electron.

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Most popular questions from this chapter

The Galileo space probe was launched on its long journey past Venus and Earth in \(1989,\) with an ultimate goal of Jupiter. Its power source is \(11.0 \mathrm{kg}\) of \(^{238} \mathrm{Pu}\) a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the \(5.59-\mathrm{MeV} \quad \alpha\) particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of \(^{238} \mathrm{Pu}\) is 87.7 years. (a) What was the original activity of the \(^{238} \mathrm{Pu}\) in becquerels? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping \(\gamma\) rays.

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Find the mass of \(^{239} \mathrm{Pu}\) that has an activity of \(1.00 \mu \mathrm{Ci}\).

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