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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minQB

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An electron is in an excited state in a hydrogen-like atom. It has a total energy of -3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is $λ$ :

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Answer:

Solution:

$K E = - E_{total} = + 3.4 eV . λ = \frac{h}{2 m K E} = \frac{12.27}{3.4 \times 1} \overset{˚}{A} \approx 6.6 \times 1 0^{- 10} m$ .

Subject:PhysicsTopic:Modern PhysicsSubtopic:De Broglie & Matter Waves

⏱ 2mℹ️ Source: QB

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