Longest wavelength of light
Quantum Theory
1. (II) (a) What is the velocity of a beam of electrons that go undeflected when passing through crossed perpendicular) electric and magnetic fields of magnitude $1.88*10^4(V/m)$ and $2.90*10^{-3} T$ respectively?
b) What is the radius of the electron orbit if the electric field is turned off?
2. (I) How hot is a metal being welded if it radiates most strongly at $440 nm$?
3. (I) What is the energy of photons (joules) emitted by an $88.5-MHz$ FM radio station?
4. (II) What is the longest wavelength of light that will emit electrons from a metal whose work function is $3.10 eV$?
5. (II) Barium has a work function of $2.48 eV$. What is the maximum kinetic energy of electrons if the metal is illuminated by UV light of wavelength $365 nm$? What is their speed?
6. (I) What is the wavelength of a neutron ($m = 1.67 x 10^{-27} kg$) traveling at $6.5 x 10^4 m/s$?
7. (I) (a) Determine the wavelength of the second Balmer line ( to transition) using Fig. 27–27. Determine likewise (b) the wavelength of the second Lyman line and (c) the wavelength of the third Balmer line.
8 (II) What is the longest wavelength of light capable of ionizing a hydrogen atom in the ground state?
You can read more on the hydrogen atom here
Answers
1. The condition of non deflection is (magnetic force = electric force)
$e*v*B = e*E$
where q is the electron charge, B and E are the magnetic and electric fields and v is the speed.
$V = E/B = 1.88*10^4/2.9*10^{-3} =6.48*10^6 (m/s)$
The radius of the circle comes from equation the centrifugal force with the magnetic force
$e*v*B = mv^2/R$ (m is the mass of electron)
$R = mv/(eB) = 9.1*10^{-31}*6.48*10^6/(1.6*10^{-19}*2.9*10^{-3}) =1.27*10^{-2} m =1.27 (cm)$
2. The Wien law for the radiated wavelength as a function of temperature is
$\lambda = b/T$ where $b$ is a constant $b =2.89*10^{-3} K*m$
hence
$T =b/\lambda = 2.89*10^{-3}/440*10^{-9} =6568 K$
3. the energy of o photon is
$E =h*F$ where h is the Planck constant and F is the frequency of the photon
$E = 6.62*10^{-34}*88.5*10^6 =5.86*10^{-26} J$
The longest wavelength of light
4. The longest wavelength of light ($\lambda$) corresponds to the energy equal to the work function.
$E=h*F =h*c/\lambda$ , $c$ is the speed of light
$E = W$
$h*c/\lambda = W$
$\lambda = h*c/W =6.62*10^{-34}*3*10^8/3.1*1.6*10^{-19} =4*10^{-7} m = 400 nm$
5. The photoelectric law is
$h*F = W+m*v^2/2$ so that $h*F = W+ Ec$
F is the frequency of light, W is the work function, m is the mass of electrons, v is the speed of electrons, Ec is the kinetic energy of electrons
$Ec = h*F-W = h*c/\lambda – W =6.62*10^{-34}*3*10^8/365*10^{-9} -2.48*1.6*10^{-19}=$
$=1.47*10^{-19} J$
The speed is
$Ec =m*v^2/2$
$v = \sqrt{2*Ec/m} =\sqrt{2*1.47*10^{-19} /9.1*10^{-31}} =5.68*10^5 m/s$
6. The de Broglie relation between wavelength associated with a particle and its impulse $p = m*v$ is
$\lambda = h/p = h/(m*v)$
$\lambda = 6.62*10^{-34}/1.67*10^{-27}*6.5*10^4 =6.1*10^{-12} m=6.1 pm$
Hydrogen transitions
7. There is no figure so I must assume the atom is Hydrogen.
The general formula for transitions from m state to n state in a hydrogen atom is
$(1/\lambda) = Rh *(1/n^2 -1/m^2)$ where m and n are the final and initial sate of the electron and $Rh$ is the Rydberg constant $Rh =10.97*10^6 m^{-1}$
For the second Balmer line $n =2$ , $m=4$
The third Balmer line has $n= 2$, $m=5$
We have for the second Lyman line $n = 1$, $m=3$
a) second Balmer ($n=2$) line
$1/\lambda = 10.97*10^6(1/4-1/16) =2.05*10^6 m$
$\lambda = 486 nm$
b) second Lyman ($n=1$) line
$1/\lambda = 10.97*10^6(1-1/9) =9.75*10^6 m$
$\lambda = 102 nm$
c) third Balmer ($n=2$) line
$1/\lambda = 10.91*10^6*(1/4-1/25) =2.29*10^6 m$
$\lambda = 436 nm$
8. The above formula for transitions applies here
$1/\lambda = Rh(1/n-1/m)$
where $n = 1$ (ground state) and $m = infinity$ (ionized atom)
hence
$1/\lambda = 10.91*10^6*(1-0) m$
$\lambda =91.7 *10^{-9} m= 91.7 nm$
