-->

Atoms

Question

The de-Broglie wavelength of electron falling on the target in an X-ray tube is),. The cut-off wavelength of the emitted X-ray is

$λ_{0} = \frac{{(m c λ)}^{2}}{h}$

$λ_{0} = \frac{m^{2} cλ}{h^{2}}$

$λ_{0} = \frac{2 mc λ^{2}}{h}$

$λ_{0} = \frac{m c λ^{2}}{h^{2}}$

Solution

$λ_{0} = \frac{2 mc λ^{2}}{h}$

The de-Broglie wavelength is given by

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}}$ .... (i)

where, E is the energy of the electron. The cut-off wavelength λ₀ is given by

$λ_{0} = \frac{h c}{E}$ ....(ii)

From Eq. (i),

λ² = $\frac{h^{2}}{2 mE}$

⇒ E = $\frac{h^{2}}{2 m λ^{2}}$ ....(iii)

Substituting the value of E from Eq. (ii), we get

$λ_{0} = \frac{h c}{\frac{h^{2}}{2 {mλ}^{2}}}$

λ₀ = $\frac{2 m {cλ}^{2}}{h}$