HOMEWORK SOLUTIONS

WEEK 3 PROBLEMS

 

Ch. 3

#15.) (a) Construct an energy level diagram for the He⁺ ion, for which Z = 2.

 

Use eq. 3.36 to determine the values of the energy levels

 

E_n = - ke² (Z²/n²)

          2a_o

 

Which, for Z = 2, gives

 

                          E_n = -54.4/n² eV

 

So

 

                          E₁ = -54.4 eV

 

                          E₂ = -13.6 eV

 

                          E₃ = -6.04 eV

 

                          E₄ = -3.4 eV

 

etc.

 

Therefore, the energy level diagram looks like

 

                       E //

                          |         etc.

                          |         _____  -6.04 eV

                          |         _____  -13.6 eV

                          |

                          |

                          |         _____  -54.4 eV

                          |___________________>

 

(b) What is the ionization energy for He⁺?

 

The ionization energy is just the energy required to allow the electron to escape from the ground state, at E₁ = -54.4 eV,

to free space, at E = 0. That means that the ionization energy is equal to the energy of the ground state

 

       E_ionization = 54.4 eV

 

 

#19.) A photon is emitted from a hydrogen atom that undergoes an electronic transition from the state n = 3 to the state n = 2. For the emitted photon, calculate

 

(a)   the energy

 

The energy of the photon is the difference between hydrogenic energy levels

 

       E_g = DE_H = E₃ – E₂

 

            = -13.6(1/9 – 1/4)

 

            = 1.89 eV

 

(b)   the wavelength

 

l = hc/E_g

 

    = 655 nm

 

(c)   the frequency

 

n = c/l

 

   = 457 x 10¹² Hz

 

 

#25.) Calculate the frequency of the photon emitted by a hydrogen atom making a transition from the n = 4 state to the n = 3 state.  Compare your result with the frequency of revolution for the electron in these two Bohr orbits.

 

This is basically the same as (a) of the last problem

 

       E_g = DE_H = E₄ – E₃

 

           = 0.661 eV

 

So that the frequency is

 

       n = E_g/h = 160 x 10¹² Hz

 

In the Bohr model, the electron in a certain energy level orbits the nucleus in a circle at a radius given by the Bohr radius

 

       r_n = n²a_o

 

So, in the n = 4 state

 

       r₄ = 8.46 x 10^-10 m

 

and the n = 3 state

 

       r₃ = 4.76 x 10^-10 m

 

The velocity of the electron in each orbit is given by the angular momentum quantization condition

 

       m_ev_nr_n = nh/(2p)

 

       v_n = nh/(2p*m_e*r_n)

 

So

 

       v₄ = 547 x 10³ m/s

 

       v₃ = 730 x 10³ m/s

 

The frequency of revolution of the electron in each state is therefore

 

       f₄ = v₄/(2pr₄) = 103 x 10¹² Hz

 

       f₃ = v₃/(2pr₃) = 244 x 10¹² Hz

 

The frequency of the emitted photon is intermediate between these two values.

 

 

 

Ch. 4

#7.) Calculate the de Broglie wavelength of a proton that is accelerated through a potential difference of 10MV

 

Given that the proton is accelerated through a potential difference of 10 MV, that means that

 

       K_p+ = 10 MeV

 

Since K_p+/m_p+c² < 0.1, we’re safe using nonrelativistic methods. So

 

       K_p+ = p²/2m

 

=> p = Ö(2m_p+K_p+)

 

The wavelength is then given by

 

       p = h/l

 

ð     l = h/p = 9.05 fm

 

 

#37.) A p^o meson is an unstable particle produced in high energy particle collisions. It has a mass-energy equivalent of about 135 MeV, and it exists for an average lifetime of only 8.7 x 10^-17 s before decaying into two gamma rays. Using the uncertainty principle, estimate the fractional uncertainty Dm/m in its mass determination.

 

The rest energy of the p^o is

 

       E = m_pc²

 

This means

 

       DE = Dm_pc²

 

The time/energy uncertainty relation reads

 

       DEDt ³ h/(2p)

 

So

 

       Dm_pc²Dt ³ h/(2p)

 

or, at minimum,

 

       Dm_p = h/(2pc²Dt)

 

So

 

Dm_p/m_p = h/(2pc²Dtm_p)

 

One only has the lifetime of the particle to measure its mass, so that the uncertainty in time is equal to the lifetime

 

       Dt = 8.7 x 10^–17 s

 

       / Dm_p/m_p = h/(4pc²Dtm_p) = 5.6 x 10^-8