The Kα X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be keV. (Round off to the nearest integer) [h=4.14 × 10-15 eV s , c=3 × 108 ms -1]

Question

The Kα X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be keV. (Round off to the nearest integer) [h=4.14 × 10-15 eV s , c=3 × 108 ms -1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Eka=Ek-EL (h c/λka)=Ek-EL EL=Ek-(h c/λka) =27.5 KeV -(12.42 × 10-7 eV m /0.071 × 10-9 m ) EL=(27.5-17.5) keV =10 keV

Q. The Kα​ X-ray of molybdenum has wavelength 0.071nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5keV, the energy of this atom when an L electron is knocked out will be___ keV. (Round off to the nearest integer) [h=4.14×10−15eVs,c=3×108ms−1]

Eka​​=Ek​−EL​ λka​​hc​=Ek​−EL​ EL​=Ek​−λka​​hc​ =27.5KeV−0.071×10−9m12.42×10−7eVm​ EL​=(27.5−17.5)keV =10keV

$E_{k_{a}} = E_{k} - E_{L}$
$\frac{h c}{λ _{k_{a}}} = E_{k} - E_{L}$
$E_{L} = E_{k} - \frac{h c}{λ _{k_{a}}}$
$= 27.5 Ke V - \frac{12.42 \times 1 0 ^{- 7} e V m}{0.071 \times 1 0 ^{- 9} m}$
$E_{L} = (27.5 - 17.5) k e V$
$= 10 k e V$