Modern Physics
Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P16EUe2[2m(UE)/h]L .

(a)

Expert Solution
Check Mark

Answer to Problem 5P

The probability of transmission P16EUe2[2m(UE)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

`    1T(E)=1+[U24E(UE)]sinh2αL        (I)

Here, T(E) is the transmission probabality and α is the wave number.

Write the expression for α .

    α=[2mh2(UE)]1/2

Here, m is the mass of proton, E is the Kinetic energy and h is the Planck’s constant.

For E<<U.

(αL)2=2mUL2h2

By hypothesis: 2mUL2h2>>1

Therefore, it can be written that sinhαL12eαL and UEU .

Using above results equation (I) becomes:

    1T(E)1+(U16E)e2αL(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

    P=(16EU)e2αLP16EUe2[2m(UE)/h]L

Conclusion:

Thus, the probability of transmission P16EUe2[2m(UE)/h]L is proved.

(b)

To determine

The numerical estimates for the exponential factor in P for each of the following cases.

(b)

Expert Solution
Check Mark

Answer to Problem 5P

The numerical estimates for the exponential factor in P in case 1 is 0.90, case 2 is 0.36, case 3 is 0.415, case 4 is 0.

Explanation of Solution

Write the expression for probability.

   P16EUe2[2m(UE)/h]L

Write the exponential factor in P

    Factor=e2[2m(UE)/h]L        (II)

Case 1:

When UE=0.01eV and L=0.1nm.

Substitute 9.11×1031kg for m , 1.055×1034Js for h, 1.60×1021J for UE and 1010m  for L in equation (II).

Factor=e2[2(9.11×1031kg)(1.60×1021J)/1.055×1034Js](1010m)=e0.10235=0.90

Case 2:

When UE=1eV and L=0.1nm.

Substitute 9.11×1031kg for m , 1.055×1034Js for h, 1.60×1019J for UE and 1010m  for L in equation (II).

Factor=e2[2(9.11×1031kg)(1.60×1919J)/1.055×1034Js](1010m)=e1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×1027kg , UE=106eV and L=1015m.

Substitute 6.7×1027kg for m , 1.055×1034Js for h, 1.60×1013J for UE and 1015m  for L in equation (II).

Factor=e2[2(6.7×1027kg)(1.60×1913J)/1.055×1034Js](1015m)=e0.878=0.415

Case 4:

When object is bowling ball whose mass is m=8.0kg , UE=1J and L=2cm.

Substitute 8.0kg for m , 1.055×1034Js for h, 1J for UE and 0.02m  for L in equation (II).

Factor=e2[2(8.0kg)(1J)/1.055×1034Js](0.02m)=e1.5×10330

Conclusion:

Thus, the numerical estimates for the exponential factor in P in case 1 is 0.90, case 2 is 0.36, case 3 is 0.415, case 4 is 0.

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