Question

Write the Bohr's postulates for hydrogen atom. How did Bohr remove the drawbacks of Rutherford's model of atom?
OR
Derive the formula for the electric field outside a charged hollow sphere with the help of Gauss's law.

Hint:

Bohr: stationary orbits, quantised angular momentum \( mvr = nh/2\pi \), and \( h\nu = E_i - E_f \); these cure Rutherford's instability and continuous-spectrum problems. For the shell, apply Gauss's law over a concentric sphere of radius \( r > R \).

Answer 1

Option 1 (Bohr's model):
Postulate 1 (Stationary orbits): The electron in a hydrogen atom revolves around the nucleus only in certain fixed circular orbits called stationary orbits. While in these orbits the electron does not radiate energy, so its total energy stays constant. The needed centripetal force is supplied by the electrostatic attraction: \( \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2} = \dfrac{mv^2}{r} \).
Postulate 2 (Quantisation of angular momentum): Only those orbits are allowed in which the angular momentum is an integral multiple of \( \dfrac{h}{2\pi} \): \( mvr = \dfrac{nh}{2\pi} \), where \( n = 1,2,3,... \) is the principal quantum number.
Postulate 3 (Transition condition): The electron emits or absorbs energy only when it jumps from one allowed orbit to another. Jumping from a higher level \( E_i \) to a lower level \( E_f \) emits a photon of frequency given by \( h\nu = E_i - E_f \).
Removing Rutherford's drawbacks:
(a) In Rutherford's model the revolving (accelerating) electron should continuously radiate, spiral inward and fall into the nucleus, making the atom unstable. Bohr's first postulate of non-radiating stationary orbits explains the stability of the atom.
(b) Rutherford's model predicted a continuous spectrum, yet atoms give a line spectrum. Bohr's third postulate of fixed energy jumps explains the observed discrete line spectrum of hydrogen.
\[\boxed{mvr = \frac{nh}{2\pi}, \qquad h\nu = E_i - E_f}\]

Option 2 (Field outside a charged hollow sphere):
Step 1: Take a hollow conducting sphere of radius \( R \) carrying total charge \( q \) spread uniformly on its surface. Find the field at an external point P at distance \( r \) from the centre, where \( r > R \).
Step 2: By spherical symmetry the field is radial and has the same magnitude \( E \) at every point of a concentric Gaussian sphere of radius \( r \).
Step 3: Gauss's law: \( \oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0} \). The charge enclosed by this Gaussian sphere is the whole charge \( q \).
Step 4: Since \( E \) is constant over the surface and parallel to \( d\vec{A} \): \( E \times 4\pi r^2 = \dfrac{q}{\varepsilon_0} \).
Step 5: Therefore \( E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2} \), directed radially outward for positive \( q \). Outside, the sphere behaves as if the whole charge were at its centre.
\[\boxed{E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}}\]