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 of the electron is E and its de Broglie wavelength is λ.

(a) E = 6.8 eV, λ ~ 6.6 x 10^-10 m

(b) E = 3.4 eV, λ ~ 6.6 x 10^-10 m 

(c) E = 3.4 eV, λ ~ 6.6 x 10^-11 m 

(d) E = 6.8 eV, λ ~ 6.6 x 10^-11 m

Answer 1

Correct Answer is: (b) E = 3.4 eV, λ ~ 6.6 x 10^-10 m 

The potential energy = -2 x kinetic energy = -2E. 

∴ total energy = -2E + E = -E = -3.4 eV

or E = 3.4 eV.

Let p = momentum and m = mass of the electron.

∴ E = p²/2m 

or p = √(2mE).

de Broglie wavelength , λ = h/p = h/√(2mE) ~ 6.6 x 10^-10 m.