Extra Credit 30

 https://phys.libretexts.org/Bookshel...ves_(Exercise)

Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

#65,  #98, #109

Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)

 https://phys.libretexts.org/Bookshel...ves_(Exercise)

#66,  #99, #110

 

Solutions 

Problem 65.

Estimate the binding energy of electrons in magnesium, given that the wavelength of 337 nm is the longest wavelength that a photon may have to eject a photoelectron from magnesium photoelectrode.

Equation for photon energy:                                                           
\begin{align*}
E=\frac{hc}{\lambda} \\
\end{align*}

and 
\begin{align*}
E={hv} \\
\end{align*}

Equation for the kinetic energy of an ejected electron: 

\begin{align*}
KE_e=hv-\Phi \\
\end{align*}

At the longest wavelength, also called cut-off wavelength, the electron is ejected with the smallest kinetic energy possible so

\begin{align*}
KE_e=0 \\
\end{align*}

After substituting for hv, the following equation is formed:

\begin{align*}
0=\frac{hc}{\lambda}-\Phi \\
\end{align*}

Solving for the binding energy: 

\begin{align*}
\Phi=\frac{hc}{\lambda} \\
\end{align*}

Plug in the given wavelength and solve:

\begin{align*}
\Phi=\frac{(6.63\times 10^{-34}Js)(3.00\times 10^8 m/s)}{337\times 10^{-9}m}
\end{align*}

\begin{align*}
\Phi=5.90\times 10^{-19}J
\end{align*}

Convert to eV: 

\begin{align*}
\Phi=5.90\times 10^{-19}J(\frac{1eV}{1.6\times 10^{-19}J})
\end{align*}

 
\begin{align*}
\Phi=3.69eV
\end{align*}

Problem 98.

Find the ionization energy of a hydrogen atom in the fourth energy state.

The equation for the energies of the electronic states of the hydrogen atom is:

\begin{align*} 
\ E_n={-R_y}{\frac{Z^2}{n^2}} \\
\ R_y=2.18\times 10^{-18} J 
\end{align*}

For hydrogen Z=1.

The initial state is 

\begin{align*}
\ n_i=4
\end{align*}

and the final state is 

\begin{align*}
\ n_f=\infty
\end{align*}

The energies for the states are then:

\begin{align*} 
\ E_4={-R_y}{\frac{1}{16}}=-1.36\times 10^{-19} J
\end{align*}

 
\begin{align*}
\ E_{\infty}={-R_y}{\frac{1}{\infty}}=0
\end{align*}

Convert to eV: 

\begin{align*} 
\ E_4=-1.36\times 10^{-19}J(\frac{1eV}{1.6\times 10^{-19}J}) \\
 =-0.85 eV
\end{align*}

To solve for the ionization energy, use the equation:

\begin{align*}
\ \Delta E=E_{n_f}-E_{n_i}
\end{align*}

Plugging in the values for the initial and final state gives:

\begin{align*}
\ \Delta E=0-(-0.85eV)
\end{align*}

\begin{align*}
\ \Delta E=0.85eV
\end{align*}

Problem 109. 

What is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?

The kinetic energy equation is:

\begin{align*} 
\ KE=\frac{mv^2}{2} \\
    =\frac{p^2}{2m}
\end{align*}

Solve for momentum:

\begin{align*} 
\ p=\sqrt{2mKE}
\end{align*}

Plug in the kinetic energy of 2.0 MeV into the equation and convert it to Joules:

\begin{align*} 
\ p=\sqrt{2(1.672\times 10^{-27}kg)(2.0\times 10^6 eV)(\frac{1.6\times 10^{-19} J}{1 eV})} \\
   =3.27\times 10^{-20} kgm/s
\end{align*}

Substitute momentum into the equation for de Broglie's wavelength:

\begin{align*} 
\ \lambda=\frac{h}{p} \\
\ =\frac{6.626\times 10^{-34} kgm^2/s}{3.27\times 10^{-20} kgm/s} \\
\ =2.03\times 10^{-14} m
\end{align*}

Do the same steps for 10.0 MeV kinetic energy:

\begin{align*} 
\ p=\sqrt{2(1.672\times 10^{-27}kg)(10.0\times 10^6 eV)(\frac{1.6\times 10^{-19} J}{1 eV})} \\
   =7.31\times 10^{-20} kgm/s
\end{align*}

 

\begin{align*} 
\ \lambda=\frac{h}{p} \\
\ =\frac{6.626\times 10^{-34} kgm^2/s}{7.31\times 10^{-20} kgm/s} \\
\ =9.06\times 10^{-15} m
\end{align*}

Problem 66.

The work function for potassium is 2.26 eV. What is the cutoff frequency when this metal is used as photoelectrode? What is the stopping potential when for the emitted electrons when this photoelectrode is exposed to radiation of frequency 1200 THz?

The cutoff frequency is:

\begin{align*} 
\ 5.45\times 10^{14} s^{-1}
\end{align*}

The stopping potential is:

\begin{align*} 
\ 2.71V
\end{align*}

Problem 99.

It has been measured that it required 0.850 eV to remove an electron from the hydrogen atom. In what state was the atom before the ionization happened?

The state it was in was n=4. 

Problem 110.

What is the de Broglie wavelength of a 10-kg football player running at a speed of 8.0 m/s?

\begin{align*} 
\lambda=8.28\times 10^{-35} m
\end{align*}