# If we assume only gravitational attraction between proton and electron in hydrogen atom and the Bohr’s quantization….

Q. If we assume only gravitational attraction between proton and electron in hydrogen atom and the Bohr’s quantization rule to be followed, then the expression for the ground state energy of the atom will be (the mass of proton is M and that of electron is m)

(a)   G2M2m2/h2

(b) −2π2G2M2m3/h2

(c) −2π2GM2m3/h2

(d) None of these

Ans: (b)

Sol: As , mv2/r = GMm/r2

Hence , v = √(GM/r)

A/c to Bohr’s postulate

mvr = nh/2π

m {√(GM/r)}r = nh/2π

Squaring ,

m2(GM/r)r2 = (nh/2π)2

For ground state putting n = 1

r = (h/2π)2 × (1/GMm2)  …..(i)

Ground State Energy ,

E = 1/2(mv2) + (-GMm/r)

= (GMm/2r) -GMm/r      (Since , v=√(GM/r)  )

= -GMm/2r

On putting the value of r from eqn(i)

E = −2π2G2M2m3/h2