The $K_{a} - X$ ray of Molybdenum has a wavelength of $71 \times 10^{- 12} m$ . If the energy of a Molybdenum atom with K-electron removed is $23.32 K e V$ , then the energy of Molybdenum atom when an L-electron removed is:
( $h c = 12.42 \times 10^{- 7} e V$ )

40.82 K e V

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23.32 K e V

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5.82 K e V

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17.5 K e V

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Solution

The correct option is C $5.82 K e V$
Given,
$h c = 12.42 \times 10^{- 7} e V$
$λ = 71 \times 10^{- 12} m$
$E_{K_{α}} = 23.32 K e V$
The formula used is:
$△ E = \frac{h c}{λ}$
Now, we have
$△ E = \frac{12.42 \times 10^{- 7} e V}{71 \times 10^{- 12} m} = 17.429 K e V$
But,
$△ E = E_{K_{α}} - E_{L}$
$E_{L} = 23.32 - 17.43 = 5.89 K e V$

Hence, the answer is OPTION C.

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Similar questions

The $K_{α}$ X-ray of molybdenum has wavelength 71 pm. If the energy of a molybdenum atom with a K electron knocked out is 23.32 keV, What will be the energy of this atom when an L electron is knocked out?

Q. The

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Q. The

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, the energy of this atom when an

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(in keV)

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[h = 4.14 \times 10^{- 15} eVs, c = 3 \times 10^{8} {ms}^{- 1}]

The $K_{α}$ and $K_{β}$ X -rays of molybdenum have wavelengths 0.71 $A^{\circ}$ and 0.63 $A^{\circ}$ respectively. Find the wavelength of $L_{α}$ X-ray of molybdenum.

Q. The K_α and K_β X-rays of molybdenum have wavelengths 0.71 A and 0.63 A respectively. Find the wavelength of L_α X-ray of molybdenum.

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The Ka−X ray of Molybdenum has a wavelength of 71×10−12m. If the energy of a Molybdenum atom with K-electron removed is 23.32 KeV, then the energy of Molybdenum atom when an L-electron removed is: (hc=12.42×10−7eV)

The correct option is C 5.82 KeVGiven,hc=12.42×10−7eVλ=71×10−12mEKα=23.32KeVThe formula used is:△E=hcλNow, we have △E=12.42×10−7eV71×10−12m=17.429KeVBut, △E=EKα−ELEL=23.32−17.43=5.89KeVHence, the answer is OPTION C.

The $K_{a} - X$ ray of Molybdenum has a wavelength of $71 \times 10^{- 12} m$ . If the energy of a Molybdenum atom with K-electron removed is $23.32 K e V$ , then the energy of Molybdenum atom when an L-electron removed is:
( $h c = 12.42 \times 10^{- 7} e V$ )