Electrons of mass with de-Broglie wavelength fall on the target in an X-ray tube. The cutoff wavelength of the emitted X-ray is [2016]

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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 [2016]

1 $λ_{0} = \frac{2 m c λ^{2}}{h}$

2 $λ_{0} = \frac{2 h}{m c}$

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

4 $λ_{0} = λ$

Ans.
(1)

Kinetic energy of electrons

$K = \frac{p^{2}}{2 m} = \frac{{(h / λ)}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$

So, maximum energy of photon (X-ray) = K

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

Solution

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 [2016]

1 $λ_{0} = \frac{2 m c λ^{2}}{h}$

2 $λ_{0} = \frac{2 h}{m c}$

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

4 $λ_{0} = λ$

Ans.
(1)

Kinetic energy of electrons

$K = \frac{p^{2}}{2 m} = \frac{{(h / λ)}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$

So, maximum energy of photon (X-ray) = K

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

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Solution

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 [2016]

1 λ0=2mcλ2h

2 λ0=2hmc

3 λ0=2m2c2λ3h2

4 λ0=λ

Ans. (1) Kinetic energy of electrons K=p22m=(h/λ)22m=h22mλ2 So, maximum energy of photon (X-ray) = K hcλ0=h22mλ2 ∴ λ0=2mcλ2h

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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 [2016]

1 $λ_{0} = \frac{2 m c λ^{2}}{h}$

2 $λ_{0} = \frac{2 h}{m c}$

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

4 $λ_{0} = λ$

Ans.
(1)

Kinetic energy of electrons

$K = \frac{p^{2}}{2 m} = \frac{{(h / λ)}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$

So, maximum energy of photon (X-ray) = K

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