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Practice Problems - Modern Physics

1. A 100-keV x ray is Compton-scattered through an angle of 90 degrees. What is the energy of the x ray after scattering?

a. 83.6 keV
b. 121 keV
c. 114.5 keV
d. 100 keV

2. What is the de Broglie wavelength of a particle moving at a speed of 1.00 x 10^6 m/s if it is (a) an electron (b) a proton? (me= 9.11 x 10^-31 kg and mp=1.67x^-27 kg.)

a. 0.727 nm and .000397 nm
b. 7.27 nm and 39.7 nm
c. 1.25 nm and .00452 nm
d. 72.7 nm and .0389 nm

3. A slowly moving electron is localized to be within 0.50mm of a particular point. What is the minimum uncertainty ith which its velocity can be simultaneously determined?

a. 23 m/s
b. 0.23 m/s
c. 46 m/s
d. 4500 m/s

4. In radiating a photon of green light (l = 546 nm), an atom radiates for 1.00 x 10^-9 s. (a) What is the energy of the photon? (b) What is the uncertainty in its energy?

a. 2.27 eV and 0.659 micro. eV
b. 8.42 eV and .01 eV
c. 1.86 eV and .42 micro. eV
d. 5.33 eV and .002 eV

5. Imagine an electron confined in one-dimensional potential box of width L = 0.12 nm. Compute the energy of the first three energy levels.

a. 3.2 x 10^(-18)J 2.7 x 10^(-17)J 4.8 x 10^(-17)J
b. 4.2 x 10^(-18)J 1.7 x 10^(-17)J 3.8 x 10^(-17)J
c. 2.2 x 10^(-18)J 1.3 x 10^(-17)J 2.8 x 10^(-17 J
d. 2.2 x 10^(-18)J 2.7 x 10^(-17)J 4.8 x 10^(-17)J

6. The ground-state energy of an electron confined along the x axis is 6.03 x 10^-18 J. What is the approximate length of the region to which the electron is confined?

a. 0.3 nm
b. 0.1 nm
c. 1.2 nm
d. 2.4 nm

7. A wave function within a barrier is described by psi(x)=psi(0)e^(-ax). What is the probability density for finding the wave at x= 3/a? Give your answer as a multiple of psi(0)^2.

a. .00248
b. 2.45
c. 1.23
d. 12.4

Solution Preview

1. wavelength of inciden photons L1 = hC/E = 1240eV nm /100*10^3eV

= 1240*10^-5 nm

Change in wavelength dL = (h/mo*C)*[1-Cos t ]

The constant in the Compton formula is given by 0.00243nm
and is called the Compton wavelength for the electron.

dL = 0.00243 * 1-0 = 0.00243 nm

L2 = L1+0.00243 = 0.0124 + 0.00243 = 0.01483 nm

New energy = hC/L2
= 1240 eV.nm/0.01483nm = 83614.3 eV = 83.6 keV --Ans A

2. deBroglie wavelength Lambda = h/p ...

Solution Summary

Notjust the answers - all questions are provided with good explanations and steps.

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