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. (answer has to be in nm)
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Physics-PLEASE HELP ASAP - Elena, Thursday, June 7, 2012 at 10:41am
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.
Elena already answered it !
it was wrong !
To find the shortest wavelength of the radiation emitted by a tungsten target when struck by accelerated electrons, we can make use of the equation known as the de Broglie wavelength.
The de Broglie wavelength (λ) of a particle can be calculated using the following equation:
λ = h / p
Where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the particle.
In this case, the accelerated electrons will have a certain momentum, and their momentum can be calculated using the equation:
p = √(2mK)
Where p is the momentum, m is the mass of the electron (9.11 x 10^-31 kg), and K is the kinetic energy of the electron.
The kinetic energy (K) of the electron can be calculated using the potential difference (V) through which the electrons are accelerated:
K = eV
Where K is the kinetic energy, e is the elementary charge (1.6 x 10^-19 C), and V is the potential difference.
Now, let's plug in the given values into the equations and solve step by step:
1. Calculate the kinetic energy (K):
K = (1.6 x 10^-19 C) × (38.9 x 10^3 V)
K = 6.224 x 10^-17 J
2. Calculate the momentum (p):
p = √(2 × (9.11 x 10^-31 kg) × (6.224 x 10^-17 J))
p = 2.73 x 10^-23 kg·m/s
3. Calculate the wavelength (λ):
λ = (6.626 x 10^-34 J·s) / (2.73 x 10^-23 kg·m/s)
λ = 2.42 x 10^-11 m
Finally, to convert the wavelength from meters to nanometers, we can multiply by 10^9:
λ (nm) = (2.42 x 10^-11 m) × (10^9 nm/m)
λ (nm) ≈ 24.2 nm
Therefore, the shortest wavelength of the radiation emitted by the tungsten target when struck by the accelerated electrons is approximately 24.2 nm.