Chapter 7: Q13P (page 310)

Find the lowest bound on the ground state of hydrogen you can get using a Gaussian trial wave function
$ψ (r) = A e^{- b r^{2}}$ ,
where A is determined by normalization and b is an adjustable parameter. Answer $- 11.5 e V$

Short Answer

Expert verified

The lowest bound on the ground state of hydrogen using Gaussian trial wave function is $<H_{m i n}> = - 11.5 e V .$

Step by step solution

Step 1: Gaussian trail wave function

A Gaussian function is proposed as a trial wave function in a variational calculation on the hydrogen atom. Determine the optimum value of the parameter and the ground state energy of the hydrogen atom. Use atomic units

$h = 2 ττ, m_{e} = 1, e = 1$

$ϕ (r, β) : = {(\frac{2 β}{π})}^{\frac{3}{4}} e x p (- β r^{2})$

Thegiventrial wave function is of the form:

$ψ (r) = A e^{- b r^{2}}$

Finding the lowest bound on the ground state of hydrogen.

First, we find the normalization constant A:

$\int ψ^{2} (r) r^{2} \sin θ d r d ϕ = 1 A^{2} \int_{0}^{\infty} e^{- 2 b r^{2}} r^{2} d r \int_{0}^{x} \sin θ d θ \int_{0}^{2 x} d ϕ = 1 A^{2} (\frac{1}{8 \sqrt{2}} \sqrt{\frac{π}{{(b)}^{3}}}) (2) (2 π) = 1 A = {(\frac{2 b}{π})}^{3 / 4}$

$<V> = - \frac{e^{2}}{4 π {\dot{0}}_{0}} {|A|}^{2} 4 π \int_{0}^{\infty} e^{- 2 {br}^{2}} \frac{1}{r} r^{2} dr = - \frac{e^{2}}{4 π {\dot{0}}_{0}} {(\frac{2 b}{π})}^{3 / 2} 4 π \frac{1}{4 b} = - \frac{e^{2}}{4 π {\dot{0}}_{0}} 2 \sqrt{\frac{2 b}{π}} .$

Step 3: Finding the value of T

Now we find <T>

$<T> = - \frac{h^{2}}{2 m} {|A|}^{2} \int e^{- b r^{2}} (\nabla^{2} e^{- b r^{2}}) r^{2} \sin θ d r d θ d ϕ B u t (\nabla^{2} e^{- b r^{2}}) = \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r} e^{- b r^{2}}) = \frac{1}{r^{2}} \frac{d}{d r} (- 2 b r^{3} e^{- b r^{2}}) = \frac{- 2 b}{r^{2}} (3 r^{2} - 2 b r^{4}) e^{- b r^{2}} = \frac{- h^{2}}{2 m} πb 4 {(\frac{2 b}{π})}^{3 / 2} (4 π) (- 2 b) \int_{0}^{\infty} (3 r^{2} - 2 {br}^{4}) e^{- 2 {br}^{2}} dr = \frac{h^{2}}{m} πb 4 {(\frac{2 b}{π})}^{3 / 2} [3 \frac{1}{8 b} \sqrt{\frac{π}{2 b}} - 2 b \frac{3}{32 b^{2}} \sqrt{\frac{π}{2 b}}] . = \frac{h^{2}}{m} 4 πb (\frac{2 b}{π}) (\frac{3}{8 b} - \frac{3}{16 b}) = \frac{3 h^{2} b}{2 m} .$

Step 4: Finding the value of H

The last two calculations of the results to get <H>

$<H> = \frac{3 h^{2} b}{2 m} - \frac{e^{2}}{4 π {\dot{o}}_{0}} 2 \sqrt{\frac{2 b}{π}}; \frac{\partial <H>}{\partial b} = \frac{3 h^{2}}{2 m} - \frac{e^{2}}{4 π {\dot{o}}_{0}} \sqrt{\frac{2}{π}} \frac{1}{\sqrt{b}} = 0 \sqrt{b} = \frac{e^{2}}{4 π {\dot{o}}_{0}} \sqrt{\frac{2}{π}} \frac{2 m}{3 h^{2}} . {<H>}_{m i n} = \frac{3 h^{2}}{2 m} {(\frac{e^{2}}{4 π {\dot{o}}_{0}})}^{2} \frac{2}{π} \frac{4 m^{2}}{9 h^{2}} - \frac{e^{2}}{4 π {\dot{o}}_{0}} 2 \sqrt{\frac{2}{π}} (\frac{e^{2}}{4 π {\dot{o}}_{0}}) \sqrt{\frac{2}{π}} \frac{2 m}{3 h^{2}} = {(\frac{e^{2}}{4 π {\dot{o}}_{0}})}^{2} \frac{m}{h^{2}} (\frac{4}{3 π} - \frac{8}{3 π}) . = - \frac{m}{2 h^{2}} {(\frac{e^{2}}{4 π {\dot{o}}_{0}})}^{2} \frac{8}{3 π} = \frac{8}{3 π} E_{1} = - 11.5 e V .$

Thus the lowest bound on the ground state of hydrogen using Gaussian trial wave function is ${<H>}_{m i n} = - 11.5 e V .$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Find the lowest bound on the ground state of hydrogen you can get using a Gaussian trial wave function
$ψ (r) = A e^{- b r^{2}}$ ,
where A is determined by normalization and b is an adjustable parameter. Answer $- 11.5 e V$

Short Answer

Step by step solution

Step 1: Gaussian trail wave function

Finding the lowest bound on the ground state of hydrogen.

Step 3: Finding the value of T

Step 4: Finding the value of H

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electromagnetism

Torque and Rotational Motion

Nuclear Physics

Electric Charge Field and Potential

Solid State Physics

Fields in Physics

Study anywhere. Anytime. Across all devices.

Company

Product

Help

Find the lowest bound on the ground state of hydrogen you can get using a Gaussian trial wave functionψ(r)=Ae-br2,where A is determined by normalization and b is an adjustable parameter. Answer-11.5eV

Short Answer

Step by step solution

Step 1: Gaussian trail wave function

Finding the lowest bound on the ground state of hydrogen.

Step 3: Finding the value of T

Step 4: Finding the value of H

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electromagnetism

Torque and Rotational Motion

Nuclear Physics

Electric Charge Field and Potential

Solid State Physics

Fields in Physics

Study anywhere. Anytime. Across all devices.

Find the lowest bound on the ground state of hydrogen you can get using a Gaussian trial wave function
$ψ (r) = A e^{- b r^{2}}$ ,
where A is determined by normalization and b is an adjustable parameter. Answer $- 11.5 e V$