The Ka X-ray emission line of tungsten occurs at I = 0.021 nm. The energy difference between K and L levels in this atoms is about:
(IIIT JEE 1997)

0.51 MeV

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1.2 MeV

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59 keV

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13.6 eV

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Solution

The correct option is C 59 keV
We know that, the energy difference is carried by the photon. So, the energy of the mentioned photon will give us the desired result.
$E = \frac{h c}{λ} = \frac{1240 e V}{0.021} = 59, 047 e V$
Which is approximately equal to 59KeV

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The Ka X-ray emission line of tungsten occurs at I = 0.021 nm. The energy difference between K and L levels in this atoms is about: (IIIT JEE 1997)

The correct option is C 59 keVWe know that, the energy difference is carried by the photon. So, the energy of the mentioned photon will give us the desired result. E=hcλ=1240eV0.021=59,047eV Which is approximately equal to 59KeV

The Ka X-ray emission line of tungsten occurs at I = 0.021 nm. The energy difference between K and L levels in this atoms is about:
(IIIT JEE 1997)

The correct option is C 59 keV
We know that, the energy difference is carried by the photon. So, the energy of the mentioned photon will give us the desired result.
$E = \frac{h c}{λ} = \frac{1240 e V}{0.021} = 59, 047 e V$
Which is approximately equal to 59KeV