Show that the probability of transmission is P ≈ 16 E U e − 2 [ 2 m ( U − E ) / h ] L .

Question

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Answer 1

Question

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/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

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.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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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Answer 2

Question

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/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

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.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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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Answer 3

Question

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/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

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.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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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Answer 4

Question

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/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

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.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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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Answer 5

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/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

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.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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.

Answer 6

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Answer 7

Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

Answer 8

(a)

Expert Solution

Answer to Problem 5P

The probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for transmission probabality.

` 1T(E)=1+[U24E(U−E)]sinh2αL (I)

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

Write the expression for α .

α=[2mh2(U−E)]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 sinh αL≈12eαL and U−E≈U .

Using above results equation (I) becomes:

1T(E)≈1+(U16E)e2αL≈(U16E)e2αL

Therefore probabality is the reciprocal of above expression.

P=(16EU)e−2αLP≈16EUe−2[2m(U−E)/h]L

Conclusion:

Thus, the probability of transmission P≈16EUe−2[2m(U−E)/h]L is proved.

Answer 13

(b)

To determine

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

(b)

Expert Solution

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.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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.

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

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.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for probability.

P≈16EUe−2[2m(U−E)/h]L

Write the exponential factor in P

Factor=e−2[2m(U−E)/h]L (II)

Case 1:

When U−E=0.01 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−21 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×10−21 J)/1.055×10−34 Js](10−10 m)=e−0.10235=0.90

Case 2:

When U−E=1 eV and L=0.1 nm.

Substitute 9.11×10−31 kg for m , 1.055×10−34 Js for h, 1.60×10−19 J for U−E and 10−10 m for L in equation (II).

Factor=e−2[2(9.11×10−31 kg)(1.60×19−19 J)/1.055×10−34 Js](10−10 m)=e−1.0235=0.36

Case 3:

When particle is alpha particles whose mass is m=6.7×10−27 kg , U−E=106 eV and L=10−15 m.

Substitute 6.7×10−27 kg for m , 1.055×10−34 Js for h, 1.60×10−13 J for U−E and 10−15 m for L in equation (II).

Factor=e−2[2(6.7×10−27 kg)(1.60×19−13 J)/1.055×10−34 Js](10−15 m)=e−0.878=0.415

Case 4:

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

Substitute 8.0 kg for m , 1.055×10−34 Js for h, 1J for U−E and 0.02 m for L in equation (II).

Factor=e−2[2(8.0 kg)(1J)/1.055×10−34 Js](0.02 m)=e−1.5×1033≈0

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.

Answer 20

Ch. 7.1 - Prob. 2E Ch. 7 - Prob. 1Q Ch. 7 - Prob. 2Q Ch. 7 - Prob. 3Q Ch. 7 - Prob. 4Q Ch. 7 - Prob. 5Q Ch. 7 - Prob. 6Q Ch. 7 - Prob. 1P Ch. 7 - Prob. 2P Ch. 7 - Prob. 3P

Show that the probability of transmission is P ≈ 16 E U e − 2 [ 2 m ( U − E ) / h ] L .

Concept explainers

Show that the probability of transmission is P≈16EUe−2[2m(U−E)/h]L .

Answer to Problem 5P

Explanation of Solution

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

Answer to Problem 5P

Explanation of Solution

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