Q26 of 29 Page 79

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 = 0.1Å, and (ii) R = 10 Å.

In this problem we will calculate ground state radius and then then treat the nucleus as point or solid charged sphere.

Let the ground state radius in normal hydrogen atom be


From Bohr equation for angular momentum we have,


--------(1)


And from force balance equation,




Now putting the equation 1 in the above equation we get,



---------(2)


On putting the values of charge and mass of electron and n=1, we get,


Case 1: since the r>R, nucleus is treated as point charge since electron will be outside the proton,




----------(3)


And


On putting the values we get, Total energy = K.E + P.E.




Case 2: since the radius r<R nucleus is treated as the solid sphere with electron orbiting inside it due to which there will be a new orbital radius of the electron at ground state r’,


Charge enclosed by the electron radius in the nucleus


Force acting on the electron will be equal to the centripetal force,




Now, after replacing r with r’ in equation 1, putting it in above equation,




Putting the value of R, we get,



Now







(using equation)


Using the formula for P.E. at point inside the solid charged sphere,




Where is the point inside the sphere and R is the radius of the sphere


on putting the values,




total energy at ground state


More from this chapter

All 29 →
24

The first four spectral lines in the Lyman series of a H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

25

Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H and 2H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common center of mass. Such a system is equivalent to a single particle with a reduced mass μ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here μ = meM/ (me + M) where M is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in 1H and 2H. (Mass of 1H nucleus is 1.6725 × 10–27 kg, Mass of 2H nucleus is 3.3374 × 10–27 kg, Mass of electron = 9.109 × 10-31 kg.)

27

In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n = 2 to n = 1 transition.

28

The inverse square law in electrostatics is for the force between an electron and a proton. The dependence of |F| can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass mp, force would be modified to where and .

Estimate the change in the ground state energy of a H-atom if mp were 10–6 times the mass of an electron.