Electrons of mass m with debroglie wavelength lambda fall on the target in an X-rays tube. The cut-off wavelength λ0 of the emitted X-rays is: (a) λ0 = λ (b) λ0 = 2mcλ2/h (c) λ0 = 2h/mc (d) λ0 = 2m2c2λ3/h2

Let Ek be the kinetic energy of the electron

Its linear momentum, p = √2mE

Using de-broglie wavelength

λ = h/p

λ = h/√2mE

The cut-off wavelength of the emitted X-rays is related to the KE of the incident electron as

E = hc/λ0

λ = h/√2m x (hc/λ0)

We get,

λ0 = 2mcλ2/h

Therefore, the correct answer is option (b)

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