The uncertainty product Δ r Δ p for the 1 s electron of a hydrogen-like atom.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

••63 SSM www In the circuit of Fig. 27-65, 8 = 1.2 kV, C = 6.5 µF, R₁ S R₂ R3 800 C H R₁ = R₂ = R3 = 0.73 MQ. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current i̟ in resistor 1, (b) current 2 in resistor 2, and (c) current i3 in resistor 3? At t = ∞o (that is, after many time constants), what are (d) i₁, (e) i₂, and (f) iz? What is the potential difference V2 across resistor 2 at (g) t = 0 and (h) t = ∞o? (i) Sketch V2 versus t between these two extreme times. Figure 27-65 Problem 63.

Thor flies by spinning his hammer really fast from a leather strap at the end of the handle, letting go, then grabbing it and having it pull him. If Thor wants to reach escape velocity (velocity needed to leave Earth’s atmosphere), he will need the linear velocity of the center of mass of the hammer to be 11,200 m/s. Thor's escape velocity is 33532.9 rad/s, the angular velocity is 8055.5 rad/s^2. While the hammer is spinning at its maximum speed what impossibly large tension does the leather strap, which the hammer is spinning by, exert when the hammer is at its lowest point? the hammer has a total mass of 20.0kg.

If the room’s radius is 16.2 m, at what minimum linear speed does Quicksilver need to run to stay on the walls without sliding down? Assume the coefficient of friction between Quicksilver and the wall is 0.236.

Answer 2

Question

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Answer 6

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Answer 7

Chapter 8, Problem 30P

To determine

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.

Answer 8

Expert Solution & Answer

Answer to Problem 30P

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

$P1s(r)=4Za03r2e−2Zr/a0$

Here, $P(r)$ is the probability density, $Z$ is the atomic number, $a0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

$〈r〉=∫0∞rP1s(r)dr$

Here, $〈r〉$ is the average distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r〉=4Za03∫0∞r3e−2Zr/a0dr$ (I)

Write the expression for Average Square of distance.

$〈r2〉=∫0∞r2P1s(r)dr$

Here, $〈r2〉$ is Average Square of distance.

Substitute $4Za03r2e−2Zr/a0$ for $P1s(r)$ in above equation.

$〈r2〉=4Za03∫0∞r4e−2Zr/a0dr$ (II)

Substitute $z$ for $2Zra0$ in equation (I).

$〈r〉=4(Za0)3(a02Z)4∫0∞z3e−zdz=14(a0Z)3!=32(a0Z)$

Substitute $z$ for $2Zra0$ in equation (II).

$〈r2〉=4(Za0)3(a02Z)5∫0∞z4e−zdz=4!8(a0Z)2=3(a0Z)2$

Write the expression for uncertainty of radius.

$Δr=(〈r2〉−〈r〉2)1/2$

Substitute $3(a0Z)2$ for $〈r2〉$ and $32(a0Z)$ for $r$ in above equation.

$Δr=(3(a0Z)2−94(a0Z)2)1/2=a0Z(3−94)1/2=0.866a0Z$

Write the expression for the average potential energy.

$〈U〉=−kZe2∫0∞1rP1s(r)dr=−4kZe2(Za0)3∫0∞re−2Zr/a0 dr=−4kZe2(Za0)3(a02Z)2=−k(Ze)2a0$

Write the expression for average momentum.

$〈p2〉=2me〈K〉=2me〈E−〈U〉〉$

Here, $〈p〉$ is the average momentum, $me$ is the mass of electron, $E$ is the total energy and $〈U〉$ is the average potential energy.

Substitute $−k(Ze)22a0$ for $E$ and $ℏ2meke2$ for $a0$ .

$〈p2〉=2mek(Ze)22a0=(Zℏa0)2$

From symmetry $〈p〉=0$ and uncertainty in momentum is:

$Δp=(〈p2〉)1/2=Zℏa0$

So product of uncertainty in momentum and uncertainty in position is:

$ΔrΔp=(0.866a0Z)(Zℏa0)=0.866ℏ$

This value $0.866ℏ$ is consistent for any value of $Z$ and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom is $0.866ℏ$ .

Answer 12

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.3 - Prob. 3E Ch. 8.5 - Prob. 4E Ch. 8 - Prob. 1Q Ch. 8 - Prob. 2Q Ch. 8 - Prob. 3Q Ch. 8 - Prob. 4Q Ch. 8 - Prob. 5Q Ch. 8 - Prob. 6Q

The uncertainty product Δ r Δ p for the 1 s electron of a hydrogen-like atom.

The uncertainty product ΔrΔp<m:math><m:mrow><m:mi>Δ</m:mi><m:mi>r</m:mi><m:mi>Δ</m:mi><m:mi>p</m:mi></m:mrow></m:math> for the 1s<m:math><m:mrow><m:mn>1</m:mn><m:mi>s</m:mi></m:mrow></m:math> electron of a hydrogen-like atom.

Answer to Problem 30P

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

The uncertainty product $ΔrΔp$ for the $1s$ electron of a hydrogen-like atom.