The wavelength of the positronium for the transition n = 3 to n = 2 .

Question

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Answer 1

Question

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

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Students have asked these similar questions

Example Two charges, one with +10 μC of charge, and another with - 7.0 μC of charge are placed in line with each other and held at a fixed distance of 0.45 m. Where can you put a 3rd charge of +5 μC, so that the net force on the 3rd charge is zero?

* Coulomb's Law Example Three charges are positioned as seen below. Charge 1 is +2.0 μC and charge 2 is +8.0μC, and charge 3 is - 6.0MC. What is the magnitude and the direction of the force on charge 2 due to charges 1 and 3? 93 kq92 F == 2 r13 = 0.090m 91 r12 = 0.12m 92 Coulomb's Constant: k = 8.99x10+9 Nm²/C² ✓

Make sure to draw a Free Body Diagram as well

Answer 2

Question

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

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Answer 3

Question

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

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Answer 4

Question

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

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Answer 5

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

Answer 6

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

Answer 7

Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

Answer 8

(a)

Expert Solution

Answer to Problem 29P

The wavelength of the positronium for the transition

$n=3$ to

$n=2$ is

$1313 nm$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

$λp=1μpZp2(36hc5k)$

$μp$ is the reduced mass for positronium,
$Zp$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2$

Substituting, $me$ for $μH$ , $(me2)$ for $μp$ , $1$ for $ZH$ , $1$ for $Zp$ , $656.3 nm$ for $λH$ to find $λp$ ,

$λp(656.3 nm)=me(1)2(me2)(1)2λp=1313 nm$

Thus, the wavelength of the positronium is $1313 nm$ .

Conclusion:

Therefore, the wavelength of the positronium is $1313 nm$ .

Answer 13

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

Answer 14

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

Answer 15

(b)

Expert Solution

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition

$n=3$ to

$n=2$ is

$164.1 nm$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula to calculate the wavelength is,

$λ=1μZ2(36hc5k)$

$μ$ is the reduced mass,
$Z$ is atomic number,
$h$ is the Planck’s constant
$c$ is the speed of light
$k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

$λHe=1μHeZHe2(36hc5k)$

$μHe$ is the reduced mass for Helium
$ZHe$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

$λH=1μHZH2(36hc5k)$

$μH$ is the reduced mass for hydrogen,
$ZH$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2$

Substituting, $me$ for $μH$ , $me$ for $μHe$ , $1$ for $ZH$ , $2$ for $Zp$ , $656.3 nm$ for $λH$ to find $λHe$ ,

$λHe(656.3 nm)=me(1)2(me)(2)2λHe=164.1 nm$

Thus, the wavelength of the Helium is $164.1 nm$ .

Conclusion:

Therefore, the wavelength of the Helium is $164.1 nm$

Answer 20

Ch. 28.3 - Prob. 28.1QQ Ch. 28.4 - Prob. 28.2QQ Ch. 28.5 - Prob. 28.3QQ Ch. 28 - Prob. 1CQ Ch. 28 - Prob. 2CQ Ch. 28 - Prob. 3CQ Ch. 28 - Prob. 4CQ Ch. 28 - Prob. 5CQ Ch. 28 - Prob. 6CQ Ch. 28 - Prob. 7CQ

The wavelength of the positronium for the transition n = 3 to n = 2 .

Concept explainers

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

Answer to Problem 29P

Explanation of Solution

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .

Answer to Problem 29P

Explanation of Solution

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Chapter 28 Solutions

The wavelength of the positronium for the transition n = 3 to n = 2 .

Concept explainers

The wavelength of the positronium for the transition n=3<m:math><m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>3</m:mn></m:mrow> </m:math> to n=2<m:math><m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>2</m:mn></m:mrow> </m:math>.

Answer to Problem 29P

Explanation of Solution

The wavelength of the singly ionized helium for the transition n=3<m:math><m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>3</m:mn></m:mrow> </m:math> to n=2<m:math><m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>2</m:mn></m:mrow> </m:math>.

Answer to Problem 29P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The wavelength of the positronium for the transition $n=3$ to $n=2$ .

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ .