To find the shortest wavelength of the radiation emitted by a tungsten target, we can use the concept of the energy of a photon. The energy of a photon is given by the equation:
E = hc/λ,
where E is the energy of the photon, h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength of the radiation.
In this case, the electrons are accelerated through a potential difference of 38.9 kV. The energy gained by an electron can be calculated by multiplying the charge of the electron (1.6 × 10^-19 C) by the potential difference (38.9 × 10^3 V):
E = qV = (1.6 × 10^-19 C) × (38.9 × 10^3 V),
where E is the energy gained by an electron.
The energy gained by an electron can also be equal to the energy of a photon emitted by the tungsten target:
E = hc/λ,
where E is the energy of the photon.
Equating the two expressions for energy, we can find the wavelength of the radiation:
hc/λ = (1.6 × 10^-19 C) × (38.9 × 10^3 V).
Now we can rearrange the equation to solve for λ:
λ = hc/[(1.6 × 10^-19 C) × (38.9 × 10^3 V)].
Plugging in the values for Planck's constant (h) and the speed of light (c), we get:
λ = (6.626 × 10^-34 J·s × 3.00 × 10^8 m/s) / [(1.6 × 10^-19 C) × (38.9 × 10^3 V)].
Calculating this expression will give us the shortest wavelength of the radiation emitted by the tungsten target.