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

Question

An electron is in an excited state in a hydrogen-like atom. It has a total energy of - 3.4 e V. The kinetic energy of the electron is E and its de Broglie wavelength is λ, E = 6 .8 eV , λ ~ 6 .6 × 10 - 10 m, E = 3 .4 eV , λ ~ 6 .6 × 10 - 10 m, E = 3 .4 eV , λ ~ 6 .6 × 10 - 11 m, E = 6 .8 eV , λ ~ 6 .6 × 10 - 11 m

MARKS App · Accepted Answer

PE = - 2 KE = - 2 E Total Energy = P E + K E = - 2 E + E = - E ⇒ - E = - 3.4 eV or E = 3.4 eV Let p is momentum and m is mass of the electron E = p 2 2 m or p = ( 2 m E ) . de Broglie wavelength λ = h p = h ( 2 m E ) λ = 6.6 x 1 0 - 10 m

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