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PhysicsModern PhysicsAtomic structureMedium2 minPYQ_2021

PhysicsMediumnumerical

The $K_{α}$ X-ray of molybdenum has wavelength $0.071 nm$ . If the energy of a molybdenum atom 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 ) $[h = 4.14 \times 10^{- 15} eV s, c = 3 \times 10^{8} {m s}^{- 1}]$

Answer:

10.00

Solution:

$E_{k_{α}} = E_{k} - E_{L}$

$\frac{h c}{λ_{k_{α}}} = E_{k} - E_{L}$

$E_{L} = E_{k} - \frac{h c}{λ_{k_{α}}}$

$= 27.5 KeV - \frac{12.42 \times 10^{- 7} eV m}{0.071 \times 10^{- 9} m}$

$E_{L} = (27.5 - 17.5) keV$

$= 10 keV$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2021

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