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For a given element the wavelength of th...

For a given element the wavelength of the `K_alpha`- line is 0.71 nm and of the `K_beta`- line it is 0.63 nm. Use this information to find wavelength of the `L_alpha` line.

Text Solution

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`E_(L_alpha) = E_(K_beta) - E_(K_alpha)`
`:. (hc)/(lambda_(L_alpha)) = (hc)/(lambda_(K_beta)) - (hc)/(lambda_(K_alpha))`
or `lambda_(L_alpha) = (lambda_(Kalpha) lambda_(K_beta))`
`=(0.71xx0.63)/(0.71 - 0.63) = 5.59 nm.`
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Knowledge Check

  • The wavelength of H_(alpha) line in Balmer series is

    A
    6653 Å
    B
    6365 Å
    C
    6563 Å
    D
    5663 Å

  • In a Coolidge tube, the potential difference used to accelerate the electrons is increased from 24. 8 kV to 49.6 kV . As a result, the difference between the wavelength of K_(alpha) -line and minimum wavelength becomes two times. The initial wavelength of the K_(alpha) line is [Take (hc)/(e) = 12 .4 kV Å ]

    A
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    B
    `(3)/(4)Å`
    C
    `(5)/(2)Å`
    D
    `(5)/(4)Å`

  • The wavelength of a spectral line is 480 nm. What is its value in mn?

    A
    `480xx10^(-7)mm`
    B
    `48xx10^(-5)mm`
    C
    `48xx10^(-6)mm`
    D
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