Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q56P

Expert-verified
Fundamentals Of Physics
Found in: Page 1305

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

An electron is emitted from a middle-mass nuclide (A=150, say) with a kinetic energy of 1.0 MeV. (a) What is its de-Broglie wavelength? (b) Calculate the radius of the emitting nucleus. (c) Can such an electron be confined as a standing wave in a “box” of such dimensions? (d) Can you use these numbers to disprove the (abandoned) argument that electrons actually exist in nuclei?

  1. The de-Broglie wavelength of the electron is 9×102fm .
  2. The radius of the emitting nucleus is 6.4 fm .
  3. No, the electron cannot be confined as a standing wave in a “box” of such dimensions due to their large wavelengths.
  4. Yes, these numbers can be used to disprove the argument that electrons actually exist.
See the step by step solution

Step by Step Solution

Step 1: The given data

  1. Mass number of the nuclide A = 150.
  2. Kinetic energy of the emitted electron.

Step 2: Understanding the concept of size and confinement

The electron can only be confined within a given structure only if its wavelength is smaller than the atomic radius of the nuclide. Now, the wavelength of an electron is associated with the momentum according to the de-Broglie concept. Thus, by comparing the wavelength with the calculated atomic radius of the nuclide, consider the strong argument for the case of confinement.

The kinetic energy of a particle in motion:

K=mc2 ….. (i)

The energy and momentum relation according to relativistic concept,

pc=K2+m2c4 …… (ii)

The atomic radius of a nuclide using its nucleon or mass number,

r=r0A13 …… (iii)

The de-Broglie wavelength of a particle of smaller size:

λ=hp …… (iv)

Step 3: a) Calculate the de-Broglie wavelength

Consider the known data, mc2=0.511MeV and hc = 1240 MeV.fm

Substitute the value of momentum from equation (ii) in equation (iv), consider de-Broglie wavelength of the electron as follows:

λ=hcK2+m2c4 =hcK2+2Kmc2 from equation (i,K=mc2)

Substitute the values as:

λ=1240 MeV.fm1.0MeV2+21.0MeV0.511MeV =9×102 fm

Hence, the value of the wavelength is 9×102 fm .

Step 4: b) Calculate the radius of the emitting nucleus

Using the given data in equation (iii), determine the radius of the emitting nucleus as follows:

r=1.2fm1501/3 =6.4fm

Hence, the value of the radius is 6.4 fm .

Step 5: c) Calculate whether the electron of the given nuclide can be confined as a standing wave

Since, λ>>r from parts (a) and (b) calculations the electron cannot be confined in the nuclide. Recall that at least λ2 is needed in any particular direction, to support a standing wave in an “infinite well.” A finite well is able to support slightly less than λ2 (as one can infer from the ground state wave function in Fig. 39-6), but in the present case λr is far too big to be supported.

Step 6: d) Determine whether an argument can be raised for part (c)

A strong cas e can be made on the basis of the remarks in part (c), above.

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.