Question

Pretend that an electron performs 1D motion in the field of the proton so that it's wavefunction is psi(x)=cos(x/a), where a=0.5 Angstrom. Remembering that the kinetic energy operator is (-hbar2/2m) d2/dx2 , compute the kinetic energy of the electron in eV (electronvolts). (I recommend comparing your answer to the energy of the 1s orbital of the hydrogen atom before you submit the answer. You should get the same order of magnitude!)

Answer 1

Wave function of the electron is

(၃) ဇ၀၁ = (2) -

Kinetic energy operator is given as

T= d.c

This kinetic energy operator operating on our given wave function will yield an eigenvalue which is the kinetic energy of the electron.

So, we have,

Tu=-20 2m d. 2 cos

2m2 cos

So, eigenvalue of the operator which is our required energy is

E = 2mg

We know mass of an electron is

m = 9.11 x 10-31 kg

And it is given that

a = 0.5 À = 0.5 x 10-10 m

Putting the values,

E (1.05 x 10-34) 2 x 9.11 x 10-31 x (0.5 x 10-102 2*

= 2.42 x 10-18 J = 2.42 x 10-18 1.6 x 10-19 ev = 15.13 eV

This is the kinetic energy of the electron.