Use the known values of ε 0 , h , m and e to calculate the following to five significant figures: h c ( in eV ⋅ nm ) , e 2 / 4 π ε 0 ( in eV ⋅ nm ) , m c 2 ( in KeV ) , a 0 ( in nm ) and E 0 ( in eV ) .

Question

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Answer 1

Question

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

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Answer 2

Question

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

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Answer 3

Question

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

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Answer 4

Question

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

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Answer 5

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

Answer 6

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

Answer 7

Chapter 4, Problem 21P

To determine

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Answer 8

Expert Solution & Answer

Answer to Problem 21P

Using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

s calculated below.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculate hc:

Here, h is the Plank’s constant and c is the speed of photon.

Substitute 4.135669×10−15 eV⋅s for h and 299792458 m/s for c in above expression.

hc=(4.135669×10−15 eV⋅s)(299792458 m/s)=1.2398×10−6 eV⋅m=1239.8 eV⋅nm

Calculate e2/ 4πε0:

Here, e is the charge in electron.

Substitute 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

e2/ 4πε0=(1.6021733×10−19 C)24π(8.8541878×10−12 F/m)(1 eV1.6021733×10−19 N⋅m )(109 nm1 m)=1.4400 eV⋅nm

Calculate mc2:

Here, m is the mass of electron.

Substitute 9.10938356×10−31kg for m and 299792458 m/s for c in above expression. mc2=(9.10938356×10−31kg) (299792458 m/s)21.602176×10−19 CeV=510999 eV=511 .00 KeV

Calculate a0:

Write the expression for a0.

a0=4πε0h2me2

Here, a0 is the Bohr’s radius, and (h) is the h - bar constant .

Substitute 9.10938356×10−31kg for m, 1.055×10−34 kg⋅m/s2 for (h), 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

a0=4π(8.8541878×10−12 F/m)(1.055×10−34 kg⋅m/s2)2(9.10938356×10−31kg)(1.6021733×10−19 C)2=5.2918×10−11 m=5.2918×10−2 nm

Calculate E0:

Write the expression for E0.

E0=e28πε0a0

Here, E0 is the Energy.

Substitute 5.2918×10−11 m for a0, 1.6021733×10−19 C for e and 8.8541878×10−12 F/m for ε0 in above expression.

E0=(1.6021733×10−19 C)28π(8.8541878×10−12 F/m)(5.2918×10−11 m)=2.179874×10−18 J(1 eV(1.6021733×10−19 J))=13.606 eV

Conclusion:

Thus, using the known values of ε0,h,m and e values of the following expressions hc is 1239.8 eV⋅nm_, e2/ 4πε0 is 1.4400 eV⋅nm_ , mc2 is 511 .00 KeV_ , a0 is 5.2918×10−2 nm_, and E0 is 13.606 eV_

Answer 12

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

Use the known values of ε 0 , h , m and e to calculate the following to five significant figures: h c ( in eV ⋅ nm ) , e 2 / 4 π ε 0 ( in eV ⋅ nm ) , m c 2 ( in KeV ) , a 0 ( in nm ) and E 0 ( in eV ) .

Use the known values of ε0,h,m and e to calculate the following to five significant figures: hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .

Answer to Problem 21P

Explanation of Solution

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Chapter 4 Solutions

Use the known values of ε0,h,m and e to calculate the following to five significant figures:

hc(in eV⋅nm ), e2/ 4πε0 (in eV⋅nm ), mc2 (in KeV ) , a0(in nm ) and E0(in eV) .