Question

The wavelength of most energetic X-rays emitted when a metal target is bombarded by 40 keV electrons, is approximately​

Answer 1

Explanation:

The wavelength of most energetic X-rays emitted when a metal target is bombarded by 40keV electrons, is approximately. (h=6.62×10-34J-sec,1eV=1.6×10-19J,c=3×108m/s).

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Answer 2

Concept:

The wavelength of rays is related to the energy and frequency by the mathematical expression, λ = hc/ eV.

Given:

Energy of electrons = 40 keV = 40 × 10³ V

Find:

We need to determine the wavelength, λ of most energetic X-rays.

Solution:

The equation eV = hv = hc/λ, where eV = energy, h = Planck's constant, = frequency, c = the speed of light, and λ = wavelength, relates energy and frequency to wavelength.

The distance between a given position and the same place in the following wave cycle is known as the wavelength.

eV = hv since v = c/λ therefore equation becomes-

eV = hc/λ

λ = hc/ eV

We know, h = planck's constant = 6.63 ×10^-34

c = 3 ×10⁸ m/s

e = 1.602 × 10^-19

Therefore, equation becomes-

λ = 6.63 ×10^-34 × 3 ×10⁸ /  1.602 × 10^-19 × 40 × 10³

λ = 19.89 × 10^-26 / 64.08 × 10^-16

λ = 0.31 × 10^-10 m

λ = 0.31 A

Thus, the wavelength of most energetic X-rays is 0.31 A.

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