An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of 45 × 10 - 2 Å . The maximum energy of a photon in the radiation in eV is h = 6.62 × 10 - 34 J s, c = 3 × 10 8 m s - 1

E = hc λ = 6.62 × 1 0 - 3 4 × 3 × 1 0 8 4 5 × 1 0 - 1 2 = 0.44 × 1 0 - 1 4 1.6 × 1 0 - 1 9 = 0.275 × 10 5 eV = 27500 eV

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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

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An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of $45 \times 10^{- 2} Å$ . The maximum energy of a photon in the radiation in $eV$ is $(h = 6.62 \times 10^{- 34} J s, c = 3 \times 10^{8} {m s}^{- 1})$

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Answer:

Solution:

$E = \frac{hc}{λ}$

$= \frac{6.62 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{45 \times 1 0^{- 12}}$

$= \frac{0.44 \times 1 0^{- 14}}{1.6 \times 1 0^{- 19}}$

$= 0.275 \times 10^{5} eV = 27500 eV$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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