Question

P3. In a hydrogen atom in its lowest energy state (known as the ground state), the electron forms a spherically-symmetric "cloud" around the nucleus, with a charge density given by ρ-A exp(-2r a ), where a,-0.529 Â-0.529 × 10-10 m is the Bohr radius. (a) Determine the constant A. (b) What is the electric field at the Bohr radius?

Answer 1

The radius of an electron orbit (spherical) in hydrogen is given by -

rn = (0.529 x 10-10 m) n2 / Z)

Z = atomic number of hydrogen atom = 1

n = principal quantum number of orbit = 1

then, we get

r = 0.529 x 10-10 m

We know that, a charge density will be given by -

= Q / V = e / [(4/3) r3)]

= (1.6 x 10-19 C) / [(4/3) (3.14) (0.529 x 10-10 m)3]

= [(1.6 x 10-19 C) / (6.19 x 10-31 m3)]

= 2.58 x 1011 C/m3

(a) The constant A which will be given as :

Here, = A exp (- 2r / a0)

(2.58 x 1011 C/m3) = A exp [- 2 (0.529 x 10-10 m) / (0.529 x 10-10 m)]

(2.58 x 1011 C/m3) = A e-2

A = [(2.58 x 1011 C/m3) / (0.135)]

A = 1.91 x 1012 C/m3

(b) The electric field at Bohr radius which will be given as -

using a formula, we have

E = k Q / a02 [(9 x 109 Nm2/C2) (1.6 x 10-19 C)] / (0.529 x 10-10 m)2

E = [(1.44 x 10-9 Nm2/C) / (2.79841 x 10-21 m2)]

E = 5.14 x 1011 N/C

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