A tungsten target is struck by electrons that have been accelerated from rest through a 38.9-kV potential difference. Find the shortest wavelength of the radiation emitted.
This question was answered several days ago by Elena and by Bob:
physics
Posted by andrej on Saturday, June 2, 2012 at 8:10am.
A tungsten target is struck by electrons that have been accelerated from rest through a 24.5-kV potential difference. Find the shortest wavelength of the radiation emitted. (in nm)
physics - bobpursley, Saturday, June 2, 2012 at 9:11am
Lets look at energy levels in the Tungsten orbitals.
Ek= Z^2*13.5eV/1^2=74^2*13.6ev/1=-74k eV
El=74^2*13.6ev/2^2=-18.6k eV
Em=74^2*13.6ev/3^2=-6.4k eV
So investigatin of what trasitions a 24.5keV electron could make, well, it cant go from k to m, but it can go from l to m.
Energy of transition: 18.6-6.4 =12.2kev
Using plancks equation;
E=hf=hc/lambda
lambda=hc/E=4.1E-15 eV s *3E8m/s *1/12.2E3 eV
lambda= 1E-10 meters=0.1 nm
check my work.
physics - Elena, Saturday, June 2, 2012 at 9:52am
For Bremsstrahlung radiation (or "braking X- radiation" )
the low-wavelength cutoff may be determined from Duane–Hunt law :
λ =h•c/e•U,
where h is Planck's constant, c is the speed of light, V is the voltage that the electrons are accelerated through, e is the elementary charge
λ = 6.63•10^-34•3•10^8/1.6•10^-19•2.45•10^4 = 5.06•10^-11 m.
physics - bobpursley, Saturday, June 2, 2012 at 10:48am
I agree with Elena on the brakding cuttoff. My answer ignores the continuous spectrum. So for the answer, consider this: What have you covered in your physics class: transitions from energy levels, or the "continuous" spectrum?
Good work, Elena.
To find the shortest wavelength of radiation emitted, we can use the equation for the energy of a photon:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the radiation.
Given that the electrons have been accelerated through a 38.9-kV potential difference, we can find the kinetic energy of the electrons. The kinetic energy (KE) of the electrons is given by:
KE = qV
where q is the charge of the electron (1.6 x 10^-19 C) and V is the potential difference.
KE = (1.6 x 10^-19 C)(38.9 x 10^3 V)
= 6.244 x 10^-17 J
Since the electrons strike the tungsten target and are emitted as radiation, the change in kinetic energy (ΔKE) of the electrons can be equated to the energy of the emitted photon:
ΔKE = hc/λ
Solving for λ, we have:
λ = hc/ΔKE
λ = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/(6.244 x 10^-17 J)
= 3.18 x 10^-10 m
Therefore, the shortest wavelength of the radiation emitted is approximately 3.18 x 10^-10 meters.
To find the shortest wavelength of the radiation emitted, we can use the energy of the accelerated electrons and the equation for the de Broglie wavelength.
The energy of an accelerated electron can be calculated using the formula:
E = qV
Where E is the energy, q is the charge of the electron (1.6 x 10^-19 C), and V is the potential difference (38.9 kV = 38.9 x 10^3 V).
Substituting the given values into the equation, we have:
E = (1.6 x 10^-19 C) x (38.9 x 10^3 V)
E ≈ 6.24 x 10^-15 J
The de Broglie wavelength of a particle is given by the equation:
λ = h / p
Where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the particle.
To find the momentum of the accelerated electron, we can use the equation:
p = √(2mE)
Where p is the momentum, m is the mass of the electron (9.11 x 10^-31 kg), and E is the energy.
Substituting the given values into the equation, we have:
p = √(2 x (9.11 x 10^-31 kg) x (6.24 x 10^-15 J))
p ≈ 8.2 x 10^-23 kg·m/s
Now, we can substitute the momentum (p) into the equation of de Broglie wavelength:
λ = (6.626 x 10^-34 J·s) / (8.2 x 10^-23 kg·m/s)
λ ≈ 8.07 x 10^-12 m
So, the shortest wavelength of the radiation emitted is approximately 8.07 x 10^-12 meters.