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The Kalpha and Kbeta X-rays of molybdenu...

The `K_alpha` and `K_beta` X-rays of molybdenum have wavelengths `0*71 A` and` 0*63 A` respectively. Find the wavelength of `L_alpha` X-ray of molybdenum.

Text Solution

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Since `K_alpha=E_K-E_L……(1) `
` lambdaK_alpha=0.71Å `
` lambdaK_beta=0.63Å `
` K_beta=E_K-E_M…….(2) `
` L_alpha=E_L-E_M…..(3) `
Subtracting (2) from (1) `K_alpha=K_beta=E_M-E_L=-L_alpha `
` L_alpha=K_beta-K_alpha `
` (3xx10^8)/(0.63xx10^-10)-(3xx10^8)/(0.71xx10^-10) `
` 4.761xx10^18-4.225xx10^18 `
` =0.526xx10^18 Hz `
Again `lambda=(3xx10^8)/(0.526xx10^-18)`
` =5.6xx10^-10=5.6Å`
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Knowledge Check

  • The K_(alpha)andK_(beta) lines of charateristic X-ray spectrum of molybdenum are 0.76 overset(@)A and 0.64overset(@)A , respectively. The wavelength of L_(alpha) line is

    A
    `1.4overset(@)A`
    B
    `2.4overset(@)A`
    C
    `4.1overset(@)A`
    D
    `3.6overset(@)A`

  • If lambda_(Cu) is the wavelength of K_(alpha) X-ray line of copper (atomic number 29 ) and lambda_(Mo) is the wavelength of the K_(alpha) X-ray line of molybdenum (atomic number 42 ),then the ratio lambda_(Cu)//lambda_(Mo) is close to

    A
    `1.99`
    B
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  • The K, L, and M energy levels of platinum lie roughly at 78, 12, and 3 KeV, respectively. The ratio of wavelength of K_(alpha) line to that of K_(beta) line in x-ray spectrum is

    A
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    C
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    D
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