An electron starting from rest and accelerated through V = 20 , 000 V in an x-ray tube loses half the energy it acquired as it approaches an atom of the target material. The remaining KE is lost in su

Question

An electron starting from rest and accelerated through $V = 20, 000 V$ in an x-ray tube loses half the energy it acquired as it approaches an atom of the target material. The remaining KE is lost in subsequent interactions with other target atoms. Find the wavelength of the radiation emitted in the first interaction.

Options:

Answer:

C

Solution:

$Δ E = \frac{1}{2} \times 2 \times 1 0^{4} = 1 \times 1 0^{4} eV = \frac{h c}{λ}$

$λ = \frac{1240}{1 0 ^{4}} eV nm = 0.124 nm = 1.24 \overset{˚}{A} .$

TestHub · Accepted Answer

Δ E = 2 1 ​ × 2 × 1 0 4 = 1 × 1 0 4 eV = λ h c ​ λ = 1 0 4 1240 ​ eV nm = 0.124 nm = 1.24 A ˚ .

Physics - Modern Physics Question with Solution | TestHub

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