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In a Coolidege tube experiment, the mini...

In a Coolidege tube experiment, the minimum wavelength of the continuouse X-ray spectrum is equal to `66.3` om . Then

A

electrons accelerate through a potential difference of `12.75 kV` in the Coolidge tube

B

electrons accelerate through a potential difference of `18.75 kV` in the Coolidge tube

C

de-Broglie wavelength of the electrons reaching the anticathode is of the order of `10 "pm"`

D

de-Broglie wavelength of the electrons reaching the anticathode is `0.01 Å`

Text Solution

Verified by Experts

The correct Answer is:
B
`V=(12400)/(lambda_(min))xx10^(-10)` volt
`V=(12400xx10^(-10))/(66.3xx10^(-12))` volt `=12.4/66.3 xx10^(4)=18.75 kV`
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Knowledge Check

  • The nature of X-rays spectrum is

    A
    continuous
    B
    line
    C
    continuous and line
    D
    None of these

  • What is the minimum wavelength of X–rays

    A
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    B
    `(hc)/(eV)`
    C
    `(hc)/(e)`
    D
    `(hc)/(V)`

  • The minimum wavelength lambda_min in the continuous spectrum of X-rays is

    A
    Proportional to the potential difference V between the cathode and anode.
    B
    Inversely proportional to potential differece V between the cathode and anode
    C
    Proportional to the square root of the potential difference V between the cathode and the anode.
    D
    Inversely proportional to the square root of the potential difference V between the cathode and the anode.

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