To show that the minimum kinetic energy is larger than 1 MeV .

Question

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Answer 1

Question

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

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Answer 2

Question

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

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Answer 3

Question

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

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Answer 4

Question

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

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Answer 5

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

Answer 6

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Answer 7

Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1 MeV.

Answer 8

(a)

Expert Solution

Answer to Problem 48E

The energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (I).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 9.1×10−31 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2=(1.055×10−34 Js)22(9.1×10−31 kg)(10−14 m)2(1 kgm2s−21 J)=6.1×10−11 J(1 eV1.6×10−19 J)=3.8125×108 eV

Thus, the energy of electron is 3.8125×108 eV which is greater than 1 MeV.

Answer 13

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

Answer 14

(b)

To determine

To see that the energy of neutron is greater than 1 MeV or not.

Answer 15

(b)

Expert Solution

Answer to Problem 48E

The energy of neutron is 2.06×105 eV which is less than 1 MeV.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

ΔxΔp≥ℏ

Here, Δx is the change in position, Δp is the change in momentum and ℏ is the reduced Planck’s constant.

Rearrange the above expression for Δp.

Δp≥ℏΔx

Write the expression for energy.

E=(Δp)22m (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute ℏΔx for Δp in equation (II).

E≥(ℏΔx)22m=ℏ22m(Δx)2

Substitute 1.055×10−34 Js for ℏ, 1.67×10−27 kg for m and 10−14 m for Δx in above equation.

E≥(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2=(1.055×10−34 Js)22(1.67×10−27 kg)(10−14 m)2(1 kgm2s−21 J)=3.3×10−14 J(1 eV1.6×10−19 J)=2.06×105 eV

Thus, the energy of neutron is 2.06×105 eV which is less than 1 MeV.

Answer 20

Ch. 27 - Prob. 1RQ Ch. 27 - Prob. 2RQ Ch. 27 - Prob. 3RQ Ch. 27 - Prob. 4RQ Ch. 27 - Prob. 5RQ Ch. 27 - Prob. 6RQ Ch. 27 - Prob. 7RQ Ch. 27 - Prob. 8RQ Ch. 27 - Prob. 9RQ Ch. 27 - Prob. 10RQ

To show that the minimum kinetic energy is larger than 1 MeV .

Concept explainers

To show that the minimum kinetic energy is larger than 1 MeV.

Answer to Problem 48E

Explanation of Solution

To see that the energy of neutron is greater than 1 MeV or not.

Answer to Problem 48E

Explanation of Solution

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Chapter 27 Solutions