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Chapter 12: Problem 18

A virus has a mass of about \(10^{-18} \mathrm{~kg}\). The de Broglie wavelength of a virus being blown in the wind at \(10 \mathrm{~m} / \mathrm{s}\) is nearest to (A) one billionth the size of a hydrogen atom. (B) one millionth the size of a hydrogen atom. (C) one thousandth the size of a hydrogen atom. (D) the size of a hydrogen atom. (E) one thousand times the size of a hydrogen atom.

Short Answer

Expert verified
(A) one thousandth (B) one millionth (C) one billionth Answer: (B) one millionth

Step by step solution

01

Know the de Broglie wavelength formula

The de Broglie wavelength formula is given by \(\lambda = \frac{h}{p}\), where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle. For an object with mass \(m\) and velocity \(v\), its momentum \(p = m \times v\). Therefore, we have \(\lambda = \frac{h}{mv}\).
02

Plug in the given data into the formula

We are given the mass \(m = 10^{-18} \mathrm{~kg}\), and the velocity \(v = 10 \mathrm{~m}/\mathrm{s}\). The Planck's constant is \(h = 6.626 \times 10^{-34} \mathrm{~Js}\). Now, substitute these values into the formula: \(\lambda = \frac{6.626 \times 10^{-34} \mathrm{~Js}}{(10^{-18} \mathrm{~kg})(10 \mathrm{~m} / \mathrm{s})}\).
03

Calculate the de Broglie wavelength

Perform the calculation from the previous step to find the de Broglie wavelength of the virus: \(\lambda = \frac{6.626 \times 10^{-34}}{10^{-17}} = 6.626 \times 10^{-17} \mathrm{~m}\).
04

Compare the de Broglie wavelength to the size of a hydrogen atom

The size of a hydrogen atom is typically around \(1 \mathrm{~Å} = 10^{-10}\mathrm{~m}\). To compare the de Broglie wavelength of the virus to the size of a hydrogen atom, divide the de Broglie wavelength by the size of a hydrogen atom, and then compare the result to the given options: \(\frac{6.626 \times 10^{-17} \mathrm{~m}}{10^{-10} \mathrm{~m}} = 6.626 \times 10^{-7}\). This value is closest to one millionth (Option B), which means that the de Broglie wavelength of the virus is about one millionth the size of a hydrogen atom. Therefore, the answer is (B).

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Most popular questions from this chapter

The half-life of the hydrogen isotope tritium is about 12 years. After a certain amount of time a fraction \(31 / 32\) of the atoms in the original sample has decayed. The time is most nearly equal to (A) 12 years (B) 24 years (C) 36 years (D) 48 years (E) 60 years

A boron atom bombarded by a slow neutron can undergo the following reaction: $$ \mathrm{n}+{ }^{10} \mathrm{~B} \rightarrow{ }^4 \mathrm{He}+{ }^7 \mathrm{Li}+\gamma . $$ The reaction products receive \(2.31 \mathrm{MeV}\) of kinetic energy. This kinetic energy should be computed as (A) equal to the kinetic energy of the neutron. (B) the total energy of the neutron. (C) the energy deficit of the reaction. (D) the energy deficit of the reaction minus the photon energy. (E) the energy deficit of the reaction minus the kinetic energy of the neutron.

The nucleus of an ordinary hydrogen atom consists of (A) a neutron. (B) a proton. (C) a proton and a neutron. (D) a proton and two neutrons. (E) two protons and two neutrons.

A hydrogen atom in its ground state absorbs a photon. In the process an electron is excited to a higher state. The atom will then most likely (A) emit a photon. (B) remain in the excited state. (C) become ionized. (D) split in half. (E) collide with another atom.

A plutonium- 239 nucleus at rest spontaneously undergoes fission via the following reaction: $$ { }_{94}^{239} \mathrm{Pu} \rightarrow{ }_{46}^{112} \mathrm{Pd}+{ }_{48}^{124} \mathrm{Cd}+3 \mathrm{n} . $$ The mass of plutonium-239 in atomic mass units is \({ }^{239} \mathrm{Pu}=239.0521\). The mass of palladium-112 is 111.9073. The mass of cadmium-124 is \(123.9176\). The mass of a neutron is \(1.0087\). a) What is the Q value of the reaction? b) If we ignore the neutrons, what should be the speeds of the palladium and cadmium nuclei? (Assume Newtonian physics.) c) If the palladium and cadmium nuclei can be treated as spheres, and are initially far apart, how close can they get to each other at these velocities? d) \({ }^{112} \mathrm{Pd}\) decays via beta emission to \({ }^{112}\) At with a half-life of 21 hrs. If a sample contains \(97 \%\) argon, how long has the palladium been decaying?
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