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Chapter 38: Q. 21 (page 1114)

The diameter of the nucleus is about 10fm. What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of10fm ?

Short Answer

Expert verified

Ek=1.3·10-12J=8.2MeV

Step by step solution

01

Given Information

The diameter of the nucleus is about 10fm. proton with a de Broglie wavelength of 10fm

02

Calculation

We can start the solution by determining de Broglie wavelength

λ=hmvv=hmλEk=mv22Ek=mh22m2λ2Ek=h22mλ2Ek=6.63·10-3422·1.67·10-27·10·10-152Ek=1.3·10-12J=8.2MeV

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Most popular questions from this chapter

A red laser with a wavelength of 650 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power. How do their rates of photon emission compare? Answer this by computing Rred/Rblue.

Electron 1 is accelerated from rest through a potential difference of 100 V. Electron 2 is accelerated from rest through a potential difference of 200 V. Afterward, which electron has the larger de Broglie wavelength? Explain

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a. Is the electron able to excite the atom? Why or why not?

b. If your answer to part a was yes, what is the electron’s kinetic energy after the collision?

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