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.
Elena and Bob replied to this question from another student:
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.
I don't know how to substitute these equations from my question please help!
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•3.89•10^4 = 3.19•10^-11 m.
To find the shortest wavelength of the radiation emitted by the tungsten target, we need to use the equation for the energy of a photon:
E = hc/λ
Where:
- E is the energy of the photon
- h is the Planck's constant (6.63 x 10^-34 J·s)
- c is the speed of light (3.00 x 10^8 m/s)
- λ is the wavelength of the radiation emitted
First, let's calculate the energy of the accelerated electrons using the potential difference:
E = qV
Where:
- E is the energy of the electrons
- q is the charge of an electron (1.60 x 10^-19 C)
- V is the potential difference (38.9 kV = 38.9 x 10^3 V)
E = (1.60 x 10^-19 C)(38.9 x 10^3 V)
Next, we can equate the energy of the accelerated electrons to the energy of the emitted photons:
E = hc/λ
Setting these two equations equal to each other and solving for λ, we get:
λ = hc/E
Substituting the values:
λ = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / [(1.60 x 10^-19 C)(38.9 x 10^3 V)]
Calculating this expression gives us the shortest wavelength of the radiation emitted by the tungsten target.