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.

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.

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 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.