An electron is in an excited state in a hydrogen like atom. It has a total energy of -3 .4 eV . The kinetic energy is E and its de Broglie wavelength is λ . Then

Question

An electron is in an excited state in a hydrogen like atom. It has a total energy of -3 .4 eV . The kinetic energy is E and its de Broglie wavelength is λ . Then, E = 6.8 eV, λ = 6.6 × 1 0 - 1 0 m, E = 3.4 eV, λ = 6.6 × 1 0 - 1 0 m, E = 3.4 eV, λ = 6.6 × 1 0 - 1 1 m, E = 6.8 eV, λ = 6.6 × 1 0 - 1 1 m

MARKS App · Accepted Answer

K . E . = K Z e 2 2 r P . E . = − K Z e 2 r T . E . = P . E . + K . E . = − K Z e 2 2 R Therefore, T E = − K E = P E 2 = − 3.4 eV So, K E = 3.14 eV Let p = momentum and m = mass of the electron. ∴ E = p 2 2 m or p = 2 mE de Broglie wavelength, λ = h p = h 2 m E On substituting the values, we get λ = 6.63 × 10 − 34 2 × 9.1 × 10 − 31 × 3.4 × 1.6 × 10 − 19 = 6.6 × 10 − 10 m

Search any question & find its solution

Looking for more such questions to practice?