Ncert Exercises & Solutions

If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R=0.1 Å and (ii) R=10 Å.

Hint: The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.

Step 1: Find the value of the Bohr radius.

\frac{{mv}^{2}}{r_{B}} = - \frac{e^{2}}{r_{B}^{2}} \cdot \frac{1}{4 {πε}_{0}}

By Bohr's postulates in the ground state, we have;

mvr = \frac{h}{2 π}

On solving,

∴ m \frac{h^{2}}{4 π^{2} m^{2} r_{B}^{2}} \cdot \frac{1}{r_{B}} = + (\frac{e^{2}}{4 {πε}_{0}}) \frac{1}{r_{B}^{2}}

∴ \frac{h^{2}}{4 π^{2} m} \cdot \frac{4 {πε}_{0}}{e^{2}} = r_{B} = 0.51 Å [This is Bohr' s radius]

Step 2: Find the potential energy and kinetic energy.

The potential energy is given by;

- (\frac{e^{2}}{4 {πε}_{0}}) \cdot \frac{1}{r_{B}} = - 27.2 eV;

KE = \frac{{mv}^{2}}{2} = \frac{1}{2} m \cdot \frac{h^{2}}{4 π^{2} m^{2} r_{B}^{2}} = \frac{h}{8 π^{2} {mr}_{B}^{2}} = + 13.6 eV

Now, for a spherical nucleus of radius R.

Step 3: Find the value of the Bohr radius for R.

If R<r_B, same result.

If R>>r_B, the electron moves inside the sphere with radius r'_B (r'_B=new Bohr radius).

Charge inside r'_{B} = e (\frac{r'_{B}^{3}}{R^{3}})

∴ r'_{B} = \frac{h^{2}}{4 π^{2} m} (\frac{4 {πε}_{0}}{e^{2}}) \frac{R^{3}}{r'_{B}^{3}}

r'_{B}^{4} = (0.51 Å) \cdot R^{3} [R = 10 Å]

= 510 {(Å)}^{4}

∴ r'_{B} ≃ {(510)}^{1 / 4} Å < R

Step 4 : Find the potential energy and kinetic energy for R .

KE = \frac{1}{2} {mv}^{2} = \frac{m}{2} \cdot \frac{h}{4 π^{2} m^{2} r'_{B}^{2}} = \frac{h}{8 π^{2} m} \cdot \frac{1}{r'_{B}^{2}}

= (\frac{h^{2}}{8 π^{2} {mr}_{B}^{2}}) \cdot (\frac{r_{B}^{2}}{r'_{B}^{2}}) = (13.6 eV) \frac{{(0.51)}^{2}}{{(510)}^{1 / 2}} = \frac{3.54}{22.6} = 0.16 eV

PE = + (\frac{e^{2}}{4 {πε}_{0}}) \cdot (\frac{r'_{B}^{2} - 3 R^{2}}{2 R^{3}})

= + (\frac{e^{2}}{4 {πε}_{0}}) \cdot (\frac{r_{B} (r'_{B}^{2} - 3 R^{2})}{R^{3}}) = + (27.2 eV) [\frac{0.51 (\sqrt{500} - 300)}{1000}]

= + (27.2 eV) \cdot \frac{- 141}{1000} = - 3.83 eV

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions