JEE Mains · Physics · STD 12 - 12. atoms
The \(K_{\alpha}\; X-\)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) \(\left[{h}=4.14 \times 10^{-15} \,{eVs}, {c}=3 \times 10^{8}\, {ms}^{-1}\right.]\)
- A \(27.5\)
- B \(17.5\)
- C \(13.6\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
As we know \({E}_{k_{a}}=E_{k}-E_{L}\) \(\frac{h c}{\lambda_{k_{a}}}=E_{k}-E_{L}\) \(E_{L}= E_{k}-\frac{h c}{\lambda_{k_{a}}}\) \(=27.5 {KeV}-\frac{12.42 \times 10^{-7}\, {eV\,m}}{0.071 \times 10^{-9}\, {m}}\) \(E_{L}=(27.5-17.5)\, {keV}\) \(=10\,{keV}\)
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