The $K_{α}$ X-Ray emission line of Tungsten occurs at 0.021 nm. Then the energy difference between K and L levels in these atoms is about

59 K e V

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29 K e V

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118 K e V

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87 K e V

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Solution

The correct option is A $59 K e V$
$λ = \frac{h c}{E_{K} - E_{L}}$
$E_{K} - E_{L} = \frac{20 \times 10^{- 26}}{0.021 \times 10^{- 9}}$
$E_{K} - E_{L} = 59 K e V$
So, the answer is option (A).

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The Kα X-Ray emission line of Tungsten occurs at 0.021 nm. Then the energy difference between K and L levels in these atoms is about

The correct option is A 59KeVλ= hcEK−ELEK−EL= 20×10−260.021×10−9EK−EL= 59 KeVSo, the answer is option (A).

The $K_{α}$ X-Ray emission line of Tungsten occurs at 0.021 nm. Then the energy difference between K and L levels in these atoms is about

The correct option is A $59 K e V$
$λ = \frac{h c}{E_{K} - E_{L}}$
$E_{K} - E_{L} = \frac{20 \times 10^{- 26}}{0.021 \times 10^{- 9}}$
$E_{K} - E_{L} = 59 K e V$
So, the answer is option (A).