Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ0) of the emitted X-ray is

Question

Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ0) of the emitted X-ray is (A) λ0 = (2 mc λ2/h) (B) λ0 = (2h/mc) (C) λ0 = (2 m2c2 λ3/h2) (D) λ0 = λ

Tardigrade Admin · Accepted Answer

beginmatrix λ0 = (hc/KEe) λ0 = (hc/h2/2m λ2) endmatrix Bigg| beginmatrix λ= (h/√2mKEe) KEe = (h2/2m λ2) endmatrix λ0 = (2mc/h) λ2

Q. Electrons of mass $m$ with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cutoff wavelength $(λ_{0})$ of the emitted X-ray is

$λ_{0} = \frac{2 m c λ ^{2}}{h}$

$λ_{0} = \frac{2 h}{m c}$

$λ_{0} = \frac{2 m ^{2} c ^{2} λ ^{3}}{h ^{2}}$

$λ_{0} = λ$

$λ_{0} = \frac{h c}{K E _{e}} λ_{0} = \frac{h c}{h ^{2} /2 m λ ^{2}} ∣ ∣ λ = \frac{h}{2 m K E _{e}} K E_{e} = \frac{h ^{2}}{2 m λ ^{2}}$
$λ_{0} = \frac{2 m c}{h} λ^{2}$

Q. Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ0​) of the emitted X-ray is

λ0​=h2mcλ2​

λ0​=mc2h​

λ0​=h22m2c2λ3​

λ0​=λ