Hydrogen Atom

Dragica VasileskaArizona State University

Importance of Hydrogen Atom

� Hydrogen is the simplest atom� The quantum numbers used to characterize the allowed

states of hydrogen can also be used to describe (approximately) the allowed states of more complex atoms� This enables us to understand the periodic table

� The hydrogen atom is an ideal system for performing precise comparisons of theory and experiment� Also for improving our understanding of atomic

structure� Much of what we know about the hydrogen atom can be

extended to other single-electron ions � For example, He+ and Li2+

Early Models of the Atom

� J.J. Thomson�s model of the atom � A volume of positive

charge� Electrons embedded

throughout the volume� A change from Newton�s

model of the atom as a tiny, hard, indestructible sphere

�watermelon� model

Experimental tests

Expect:

1. Mostly small angle scattering

2. No backward scattering events

Results:

1. Mostly small scattering events

2. Several backward scatterings!!!

Early Models of the Atom

� Rutherford�s model

� Planetary model� Based on results of

thin foil experiments� Positive charge is

concentrated in the center of the atom, called the nucleus

� Electrons orbit the nucleus like planets orbit the sun

Problem: Rutherford�s model

The �size� of the atom in Rutherford�s model is about 1.0 × 10�10 m. (a) Determine the attractive electrical force between an electron and a proton separated by this distance. (b) Determine (in eV) the electrical potential energy of the atom.

The �size� of the atom in Rutherford�s model is about 1.0 × 10�10 m. (a) Determine the attractive electrical force between an electron and a proton separated by this distance. (b) Determine (in eV) the electrical potential energy of the atom.

Given:

r = 1.0 × 10�10 m

Find:

(a) F = ?(b) PE = ?

Electron and proton interact via the Coulomb force

29 2 2 19

1 222 10

8

8.99 10 1.60 10

1.0 10

2.3 10

e

N m C Cq qF k

r m

N

Potential energy is

181 219

12.3 10 14 eV

1.6 10e

q q eVPE k J

r J

Difficulties with the Rutherford Model

� Atoms emit certain discrete characteristic frequencies of electromagnetic radiation� The Rutherford model is unable to explain this

phenomena

� Rutherford�s electrons are undergoing a centripetal acceleration and so should radiate electromagnetic waves of the same frequency� The radius should steadily decrease as this radiation

is given off� The electron should eventually spiral into the nucleus� It doesn�t

Emission Spectra� A gas at low pressure has a voltage applied to it� A gas emits light characteristic of the gas� When the emitted light is analyzed with a spectrometer, a series of

discrete bright lines is observed� Each line has a different wavelength and color� This series of lines is called an emission spectrum

Emission Spectrum of Hydrogen

� The wavelengths of hydrogen�s spectral lines can be found from

� RH is the Rydberg constant� RH = 1.0973732 x 107 m-1

� n is an integer, n = 1, 2, 3, �� The spectral lines correspond to

different values of n� A.k.a. Balmer series� Examples of spectral lines

� n = 3, ë = 656.3 nm� n = 4, ë = 486.1 nm

22H n1

21

R1

Absorption Spectra

� An element can also absorb light at specific wavelengths� An absorption spectrum can be obtained by passing a

continuous radiation spectrum through a vapor of the gas

� The absorption spectrum consists of a series of dark lines superimposed on the otherwise continuous spectrum� The dark lines of the absorption spectrum coincide

with the bright lines of the emission spectrum

Applications of Absorption Spectrum

� The continuous spectrum emitted by the Sun passes through the cooler gases of the Sun�s atmosphere� The various absorption lines can be used to identify

elements in the solar atmosphere� Led to the discovery of helium

Recall Bohr�s Assumptions

� Only certain electron orbits are stable. Radiation is emitted by the atom when the electron �jumps� from a more energetic initial state to a lower state

� The size of the allowed electron orbits is determined by a condition imposed on the electron�s orbital angular momentum

i fE E hf

, 1, 2,3,...em vr n n

Why is that?

Modifications of the Bohr Theory � Elliptical Orbits

� Sommerfeld extended the results to include elliptical orbits� Retained the principle quantum number, n� Added the orbital quantum number, ℓ� ℓ ranges from 0 to n-1 in integer steps

� All states with the same principle quantum number are said to form a shell

� The states with given values of n and ℓ are said to form a subshell

Modifications of the Bohr Theory �Zeeman Effect and fine structure

� Another modification was needed to account for the Zeeman effect� The Zeeman effect is the splitting of spectral lines in a strong

magnetic field� This indicates that the energy of an electron is slightly modified

when the atom is immersed in a magnetic field� A new quantum number, m ℓ, called the orbital magnetic quantum

number, had to be introduced� m ℓ can vary from - ℓ to + ℓ in integer steps

� High resolution spectrometers show that spectral lines are, in fact, two very closely spaced lines, even in the absence of a magnetic field� This splitting is called fine structure� Another quantum number, ms, called the spin magnetic quantum

number, was introduced to explain the fine structure

de Broglie Waves

� One of Bohr�s postulates was the angular momentum of the electron is quantized, but there was no explanation why the restriction occurred

� de Broglie assumed that the electron orbit would be stable only if it contained an integral number of electron wavelengths

de Broglie Waves in the Hydrogen Atom

� In this example, three complete wavelengths are contained in the circumference of the orbit

� In general, the circumference must equal some integer number of wavelengths

2 , 1,2,3,...r n

e

h

m v

, 1, 2,3,...em vr n n

This was the first convincing argument that the wave nature of This was the first convincing argument that the wave nature of matter was at the heart of the behavior of atomic systemsmatter was at the heart of the behavior of atomic systems

Quantum Mechanics and the Hydrogen Atom

� One of the first great achievements of quantum mechanics was the solution of the wave equation for the hydrogen atom

� The significance of quantum mechanics is that the quantum numbers and the restrictions placed on their values arise directly from the mathematics and not from any assumptions made to make the theory agree with experiments

Problem: wavelength of the electron

Determine the wavelength of an electron in Determine the wavelength of an electron in the third excited orbit of the hydrogen the third excited orbit of the hydrogen atom, with atom, with n n = 4.= 4.

Determine the wavelength of an electron in the third excited orbit of the hydrogen atom, with n = 4.

Given:

n = 4

Find:

e = ?

Recall that de Broglie�s wavelength of electron depends on its momentum, = h/(mev). Let us find it,

,e n en

nm vr n so m v

r

Recall that

20

0

,2n e

hr n a so m v

a n

Thus, 02 8 0.0529 1.33e

ha n nm nm

m v

Quantum Number Summary

� The values of n can increase from 1 in integer steps� The values of ℓ can range from 0 to n-1 in integer steps� The values of m ℓ can range from -ℓ to ℓ in integer steps

Spin Magnetic Quantum Number

� It is convenient to think of the electron as spinning on its axis� The electron is not

physically spinning� There are two directions

for the spin� Spin up, ms = ½� Spin down, ms = -½

� There is a slight energy difference between the two spins and this accounts for the Zeeman effect

Electron Clouds

� The graph shows the solution to the wave equation for hydrogen in the ground state� The curve peaks at the

Bohr radius� The electron is not

confined to a particular orbital distance from the nucleus

� The probability of finding the electron at the Bohr radius is a maximum

Electron Clouds

� The wave function for hydrogen in the ground state is symmetric� The electron can be

found in a spherical region surrounding the nucleus

� The result is interpreted by viewing the electron as a cloud surrounding the nucleus� The densest regions of

the cloud represent the highest probability for finding the electron

Roadmap for solution of Hydrogen-like atoms

� Start from 3D TISE for electron in Coulomb potential of nucleus

� Separate variables to give 1D radial problem and angular problemSolution of angular part already knownin terms of spherical harmonics.

� Simplify 1D radial problem using substitutions and atomic units

� Solve radial problemExtract asymptotic solution at large rUse Frobenius methodFind eigenvalues by requiring normalizable solutions

Mathematical Details

Reminder: SE in three dimensions

Wavefunction and potential energy are now functions of three spatial coordinates:

Kinetic energy involves threecomponents of momentum

Interpretation of wavefunction:

, ,

, ,

x x y z

V x V V x y z

r

r

2 2 22 2

2 2 2 2 2 2 22

2 2 2 2

2 2 2

2 2 2

x y zxp p pp

m m m

m x m m x y z

p

23 ,d tr r

probability of finding particle in a volume element centred on r

2

, t r

probability density at r(probability per unit volume)

2

2

2V E

m r r r r

Time-independent Schrödinger equation

H-atom is our first example of the 3D Schrödinger equation

�H E r r r

The Hamiltonian for a hydrogenic atom

2 22

0

�2 4e

ZeH

m r

In a hydrogenic atom or ion with nuclear charge +Ze there is the Coulomb attraction between electron and nucleus. This has spherical symmetry � potential only depends on r. This is known as a CENTRAL POTENTIAL

2

0

( )4

ZeV r

r

The Hamiltonian operator is

; electron mass, nuclear masse Ne e N

e N

m mm m m

m m

NB: for greater accuracy we should use the reduced mass not the electron mass here. This accounts for the relative motion of the electron and the nucleus (since the nucleus does not remain precisely fixed):

r

+Ze

-e

Hamiltonian for hydrogenic atomsThe natural coordinate system is spherical polars. In this case the Laplacian operator becomes (see 2B72):

2 2 22

2 20

�( ) ( )( ) ( )

2 2 4

L Zer E

mr r r mr r

r rr r

22 2

2 2

1 1� sinsin sin

L

22 2

2 2 2

1 Lr

r r r r

TISE for H-like atom is(m = me from now on) �H E r r r

2 2 2 2 2

2 22 2

0 0

��

2 4 2 2 4e e e

Ze L ZeH r

m r m r r r m r r

r

So the Hamiltonian is

So we will look for a solution of the form

CONCLUSIONS

� The angular momentum about any axis and the total angular momentum commute with the Hamiltonian

� They are therefore both conserved quantities

� We can have simultaneous eigenfunctions of these operators and the Hamiltonian

� We can have well-defined values of these quantities and the energy at the same time

2

2�d

� �[ , ] 0 0d�d

� �[ , ] 0 0d

z

z

LH L

t

LH L

t

Angular momentum and the H atom

2 2 2

22 2

0

��

2 2 4

L ZeH r

mr r r mr r

r

Reminder: Ehrenfest�s theorem

�d �� ,d

Q i QH Q

t t

The angular wavefunction

The angular part are the eigenfunctions of the total angular momentum operator L2. These are the spherical harmonics, so we already know the corresponding eigenvalues and eigenfunctions (see §5):

Note: this argument works for any spherically-symmetric potential V(r), not just the Coulomb potential.

2 2

� , ,

� , 1 ,

z lm lm

lm lm

L Y m Y

L Y l l Y

00

11

10

1 1

1( , )

4

3( , ) sin exp( )

8

3( , ) cos

4

3( , ) sin exp( )

8

Y

Y i

Y

Y i

2 2�Eigenvalues of are ( 1) , with 0,1,2,

�Eigenvalues of are , with z

L l l l

L m l m l

l = principal angular momentum quantum number.

m = magnetic quantum number (2l+1 possible values).

( ) ( ) ( ) ( )lmr R r Y rThis suggests we look for separated solutions of the form

The radial equationSubstitute separated solution into the time-independent Schrödinger equation

( , , ) ( ) ( , )lmr R r Y 2 2 22

2 20

�( ) ( )

2 2 4

L Zer E

mr r r mr r

r r

2 2 22

2 20

( 1)

2 2 4

d dR l l Zer R R ER

mr dr dr mr r

Get radial equation

Note that this depends on l but not on m: R(r) and E therefore involve the magnitude of the angular momentum but not its orientation.

The radial equation (2)

Define a new radial function ÷(r) by:

( )( )

rR r

r

2 2 22

2 20

( 1)

2 2 4

d dR l l Zer R R ER

mr dr dr mr r

2 2 2 2

2 20

( 1)

2 2 4

d l l ZeE

m dr mr r

Get radial equation for ÷(r)

The effective potentialNew radial equation looks like the 1D Schrödinger equation with an effective potential

2 2

eff 20

( 1)( )

2 4

l l ZeV r

mr r

r

V(r)

2 2 2 2

2 20

( 1)

2 2 4

d l l ZeE

m dr mr r

2 2

eff2( )

2

dV r E

m dr

2

2

( 1)

2

l l

mr

is known as the centrifugal barrier potential

2 2

2( )

2

dV x E

m dx

1D TISE

Where does the centrifugal barrier come from?

L mv r

2 2

3

mv LF

r mr

2

22

dV LF V

dr mr

Alternatively, we can say that the energy required to supply the extraangular speed must come from the radial motion so this decreases asif a corresponding outward force was being applied.

For circular motion

CLASSICAL ARGUMENT

Fixed l corresponds to fixed angular momentum for the electron.so as r becomes small, must increase in order to maintain L. This causes an increase in the apparent outward force (the �centrifugal� force).

L mv rv

The centrifugal barrier2

2

( 1)( )

2cb

l lV r

mr

� Simplify equation using atomic units

� Solve equation in the asymptotic limit (large r)Gives a decaying exponential solution

� Define new radial function by factoring out asymptotic solution

� Solve equation for F(r) using the series (Frobenius) method

� Find that solution is not normalizable unless series terminates.This only happens if the eigenvalues have certain special values.Hence we find the eigenvalues and eigenstates of the H-atom.

Roadmap for solution of radial equation

( ) ( ) exp( )r F r r

( ) exp( )r

r r

( ) p sp

p

F r a r

2 2 2 2

2 20

( 1)

2 2 4

d l l ZeE

m dr mr r

( ) ( ) ( )

( )( )

nlm lmR r Y

rR r

r

r

Atomic unitsThere are a lot of physical constants in these expressions. It makes atomic problems simpler to adopt a system of units in which as many as possible of these constants are one. In atomic units we set:

2 1

23 2

0

20

2

Planck constant 1 (dimensions [ ])

Electron mass 1 (dimensions [ ])

Constant apearing in Coulomb's law 1 (dimensions [ ])4

It follows that:

4 Unit of length = 5.2917

e

e

ML T

m M

eML T

e m

110

2218

20

7 10 m=Bohr radius,

Unit of energy = 4.35974 10 J 27.21159eV=Hartree, 4

eh

a

meE

In these units the radial equation becomes

2 2 2 2 2

2 2 2 20

( 1) 1 ( 1)

2 2 4 2 2

d l l Ze d l l ZE E

m dr mr r dr r r

Asymptotic solution of radial equation (large r)

( ) ( ) exp( )r F r r

Consider the radial equation at very large distances from the nucleus, when both terms in the effective potential can be neglected. We are looking for bound states of the atom where the electron does not have enough energy to escape to infinity (i.e. E < 0):

Inspired by this, rewrite the solution in terms of yet another unknown function, F(r):

2

Put 2

E

2

2 2

1 ( 1)

2 2

d l l ZE

dr r r

22

2( ) exp( )

dr r

dr

For normalizable solutions we must take the decaying solution

This gives

Differential equation for F

Derive equation for F

( ) ( ) exp( )r F r r

2

2 2

1 ( 1)

2 2

d l l ZE

dr r r

2

2 2

( 1) 22 0

d d l l ZF r

dr dr r r

Differential equation for F:

Series solution (1)Look for a power-series solution (Frobenius method). The point r = 0 is a regular singular point of the equation so at least one well-behaved series solution should exist (see 2B72).

( ) p sp

p

F r a r

2

2 2

( 1) 22 0

d d l l ZF r

dr dr r r

2 1

0 0

1 1 2 2 0p s p sp p

p p

a r p s p s l l a r p s Z

Substitute

Series solution (2)The indicial equation that fixes s comes from equating coefficients of the lowest power of r which is s - 2

We need the regular solution that will be well-behaved as r→0, so take

2 1

0 0

1 1 2 2 0p s p sp p

p p

a r p s p s l l a r p s Z

2

( 1) ( 1) 0

( 1) 0

( 1) 0

s s l l

s s l l

s l s l

, or 1s l s l

1s l

Series solution (3)

1

0 0

1 1 2 1 2 0p l p lp p

p p

a r p l p l l l a r p l Z

1 2 1

2 1 1p

p

p l Za

a p l p l l l

General recursion relation comes from equating coefficients of r to the power p+l

Series solution (4)

0 0

11

2 22

1 11

(2 ) (2 )exp(2 ) with .

! !

Coefficient of would be .

Coefft. of (2 ) ( 1)! 2 2

Coefft. of (2 ) ( 2)! ( 2)

n nn

n nn n

p lp l

p l p lp l

pp l p lp l

rr b r b

n n

r b

br p l

r b p l p l p

For p→∞ we find:

Compare with: 1(remember is coefficient of in the expansion)p lpa r

So, our series behaves for large p just like exp(2êr).

1 2p

pp

a

a p

1 2 1

2 1 1p

p

p l Za

a p l p l l l

Series solution (5)

( 1) for some integer 0,1,2

where is a positive integer : 1, 2

Zp l p

n n l n l l

To prevent this the series must terminate after a finite number of terms. This only happens if

So finally the energy is

So, if the series continues to arbitrarily large p, the overall solution becomes

( ) exp(2 ) exp( ) exp( )r r r r

n is known as the principal quantum number.It defines the �shell structure� of the atom.

(not normalizable)

2 2

22 2

ZE

n

( ) ( ) exp( )r F r r

with n > l

/

( ) ( ) ( )

( )( )

( )( )

nlm nl lm

nllm

Zr nnl

lm

R r Y

rY

r

F r eY

r

r

1

0

( ) l pnl p

p

F r r a r

1 2 1

2 1 1p

p

Z p l na

a n p l p l l l

2

2, 1, 2,3

2n

ZE n n l

n

Each solution of the time-independent Schrödinger equation is defined by three quantum numbers n,l,m

Summary of solution so far

The energy only depends on the principal quantum number n which is bigger than l

The radial solution depends on n and l but not m

1 0

2 0,1

3 0,1, 2

n l

n l

n l

2

1

2nEn

In eV ground state energy= -13.6eV= - ionisation energy

Traditional spectroscopic nomenclature:l = 0: s states (from �sharp� spectral lines)l = 1: p states (�principal�)l = 2: d states (�diffuse�)l = 3: f states (�fine�)�and so on alphabetically (g,h,i� etc)

The hydrogen energy spectrum

n > l so

In Hartrees 27.2eVhE

This simple formula agrees with observed spectral line frequencies to within 6 parts in ten thousand

The energy spectrum: degeneracy

2

2 (in atomic units)

2n

ZE

n

, ( 1), 0, ( 1),m l l l l

12

0

(2 1) .n

l

l n

For each value of n = 1,2,3� we have a definite energy:

For each value of n, we can have npossible values of the total angular momentum quantum number l:

For each value of l and n we can have 2l+1 values of the magnetic quantum number m:

l = 0, 1, 2,�, n-1

The total number of states (statistical weight) associated with a given energy En is therefore: (This neglects electron spin. See Section 7.)

The fact that the energy is independent of m is a feature of all spherically symmetric systems and hence of all atoms. The independence on l is a special feature of the Coulomb potential, and hence just of hydrogenic atoms. This is known as accidental degeneracy.

The radial wavefunctions

3/ 2

10 00

3/ 2

210 0 0

3/ 2

200 0 0

3/ 2 2

320 0 0

3/

310

( ) 2 exp( / )

1( ) exp

2 23

( ) 2 1 exp2 2 2

4( ) exp

3 327 10

4 2( )

9 3

ZR r Zr a

a

Z Zr ZrR r

a a a

Z Zr ZrR r

a a a

Z Zr ZrR r

a a a

ZR r

a

2

0 0 0

3/ 2 2 2

30 20 0 0 0

1 exp6 3

2 2( ) 2 1 exp

3 3 27 3

Zr Zr Zr

a a a

Z Zr Z r ZrR r

a a a a

Rnl(r) depends on n and l but not on m

/

( ) ( ) ( )

( ) ( )( )

nlm nl lm

Zr nnl nl

nl

R r Y

r F r eR r

r r

r

For atomic units set a0 = 1

The radial wavefunctions (2)

Full wavefunctions are:

Normalization chosen so that:

Only s states (l = 0) are finite at the origin.

Radial functions have (n-l-1) zeros (excluding r = 0).

( ) ( ) ( )nlm nl lmR r Y r

Asymptotic solution

1( ) exp( / )nnl r

R r r Zr n

Solution near r = 0

0( ) l

nl rR r r

2 2

01

nldrr R r

3/ 2

10 00

3/ 2

210 0 0

3/ 2

200 0 0

3/ 2 2

320 0 0

3/

310

( ) 2 exp( / )

1( ) exp

2 23

( ) 2 1 exp2 2 2

4( ) exp

3 327 10

4 2( )

9 3

ZR r Zr a

a

Z Zr ZrR r

a a a

Z Zr ZrR r

a a a

Z Zr ZrR r

a a a

ZR r

a

2

0 0 0

3/ 2 2 2

30 20 0 0 0

1 exp6 3

2 2( ) 2 1 exp

3 3 27 3

Zr Zr Zr

a a a

Z Zr Z r ZrR r

a a a a

Total probability density 22 2| ( ) | ( ) ( )nlm nl lmR r Y r

22 2 2 2

,

( ) ( ) ( )nl nl lmr R r dr R r Y r drd

= probability of finding particle in avolume element centred on (r, è, ö)

Radial probability density

= probability of finding the particle in aspherical shell centred on r, i.e. at any angle

3 2d r drd r sind d d

2 3| ( ) |nlm d r r

2 2 ( )nlr R r dr

(solid angle element)

Integrate over all angles using normalization of spherical harmonics

2 2 2( ) ( )nl nlr R r dr r dr so this is analogous to the 1D case

(Rnl is real)

Radial probability density

Total probability density

2 3 2 2 2| ( ) | ( ) . | ( ) | sinnlm nl lmd r R r dr Y d d r r

Solid angle probability density

= probability density of finding particle in a solid angle element sind d d

2| ( ) |lmY

2

2

| ( ) |

| ( ) | sin

lm

lm

Y d

Y d d

= probability of finding particle betweenè and è + dè and ö and ö + dö

= (Radial probability) x (Angular probability)

Angular probability density

3 *

23

0

( ) ( )nlm nlm

nl

r d r

dr r R r

r r r

13.53 03s

5.02 12p

6.02 02s

1.51 01s

<r> (au)n lOrbital

2 2 ( )nlr R r

Radial probability density

Radial probability density

Comparison with Bohr model

, 1, 2,3,zL n n

20

0, Bohr radiusn

n ar a

Z

Angular momentum (about any axis) assumed to be quantized in units of Planck�s constant:

Electron otherwise moves according to classical mechanics and has a single well-defined orbit with radius

Energy quantized and determined solely by angular momentum:

Bohr model Quantum mechanics

2

2, Hartree

2n h h

ZE E E

n

, , ,zL m m l l

020

1 1 , Bohr radiusn

Zar rn a

2

2, Hartree

2n h h

ZE E E

n

Angular momentum (about any axis) shown to be quantized in units of Planck�s constant:

Energy quantized, but determined solely by principal quantum number n, not by angular momentum:

Electron wavefunction spread over all radii. Can show that the quantum mechanical expectation value of 1/r satisfies

These results are not exact because we have made several approximations.

� We have neglected the motion of the nucleus. To fix this we should replace meby the reduced mass ì. This improves agreement with experiment by an order of magnitude (simple formula gives spectral lines to within 4 parts in 100 thousand!)

� We have used a non-relativistic treatment of the electron and in particular have neglected its spin (see §7). Including these effects give rise to

�fine structure�(from the interaction of the electron�s orbital motion with its spin)

�hyperfine structure�(from the interaction of the electron�s spin with the nuclear spin)

� We have neglected the fact that the EM field between the nucleus and the electron is itself a quantum object. This leads to �quantum electrodynamic�(QED) corrections, and in particular to a small �Lamb shift� of the energy levels.

The remaining approximations

2

2

and ( )

1 1( ) ( )

1 / 1 / 2

e pe e

e p

he p e p

m mm E n m

m m

ZE n E n E

m m m m n

Ground state energy = -1/2 au = -13.6eV = - ionisation energy

1

2

0

2 1n

l

g l n

Statistical weight

2 2 ( )nlr R rRadial probability density

Energy levels in au

Wavefunction

Summary

/

( ) ( ) ( )

( )( )

nlm nl lm

Zr nnl

lm

R r Y

F r eY

r

r1

0

( ) l pnl p

p

F r r a r

1 2 1

2 1 1p

p

Z p l na

a n p l p l l l

2

2, 1, 2,3

2n

ZE n n l

n

In Hartrees 27.2eVhE

Permissible Quantum States

Orbital energies of the hydrogen atom.

Orbital Energies of MultielectronAtoms

� All elements have the same number of orbitals (s,p, d, and etc.).

� In hydrogen these orbitals all have the same energy.

� In other elements there are slight orbital energy differences as a result of the presence of other electrons in the atom.

� The presence of more than one electron changes the energy of the electron orbitals(click here)

Shape of 1s Orbital

Shape of 2p Orbital

Shape of 3d Orbitals

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