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)
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.
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.
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 the radiation emitted when electrons strike a tungsten target, we can use the equation relating the energy of a single photon to its wavelength:
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 (3.0 x 10^8 m/s),
and λ is the wavelength of the light.
To determine the energy of the photons, we need to know the kinetic energy of the electrons before they strike the target. This can be calculated using the potential difference through which the electrons are accelerated.
The formula for the kinetic energy of an electron is given by:
KE = eV
Where:
KE is the kinetic energy,
e is the elementary charge (1.602 x 10^-19 C),
and V is the potential difference.
Given a potential difference of 24.5 kV, we can convert it to volts by multiplying by 1000:
V = 24.5 kV = 24.5 x 1000 V = 24,500 V
Now we can calculate the kinetic energy of the electrons:
KE = eV = (1.602 x 10^-19 C) (24,500 V) = 3.93 x 10^-15 J
Next, we can use the kinetic energy of the electrons to find the energy of each photon emitted by the electrons when they strike the tungsten target:
E = KE
Now we can rearrange the equation to solve for the wavelength:
λ = hc/E
Substituting the known values:
λ = (6.626 x 10^-34 J·s) (3.0 x 10^8 m/s) / (3.93 x 10^-15 J)
Simplifying, we get:
λ = (1.988 x 10^-8 m) / (3.93 x 10^-15 J)
λ ≈ 5.06 x 10^-7 m
Finally, to convert the wavelength from meters (m) to nanometers (nm), we can multiply the value by 10^9:
λ ≈ 5.06 x 10^-7 m x 10^9 nm/m
λ ≈ 506 nm
Therefore, the shortest wavelength of the radiation emitted when electrons strike the tungsten target is approximately 506 nm.