A tungsten target is struck by electrons that have been accelerated from rest through a 40KV potential difference. Find the shortest wavelength of the bremsstrahlung radiation emitted? (Ans: 0.31 x 10⁻¹⁰ m)
Given:
Voltage = V = 40 kV = 40 x 10³ V
∴ Plank's Constant = h = 6.62 x 10⁻³⁴ m² kg s⁻¹
∴ Charge on the electron = e = 1.6 x 10⁻¹⁹ C
∴ Velocity of light = c = 3 x 10⁸ m s⁻¹
To Find:
Shortest Wavelength = λmin= ?
Solution:
We know that (from v=fλ)
fmax = cλmin
or
λmin =cfmax
where Since Emax = h fmax ⇒ fmax = Emaxh and where Emax = K.E. = Ve (according to law of conservation of energy) So,
λmin =hcVe
by putting values
λmin =`\frac {6.62 x 10⁻³⁴ m² kg s⁻¹ x 3 x 10⁸ m s⁻¹}{40 x 10³ V x 1.6 x 10⁻¹⁹ C }`
λmin = 19.86x10⁻²⁶m³kgs⁻²64x10⁻¹⁶ VC
λmin = 0.3103125 x 10⁻¹⁰ V
λmin = 0.31 x 10⁻¹⁰ V --------------Ans.
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