Question:

Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ( 10^−10m). (a) Construct a quantity with the dimensions of length from the fundamental constants e,me​,andc. Determine its numerical value.

Answer:

a) Construct a quantity with the dimensions of length from the fundamental constants e, me and c:

Length = (e^2)/(me*c^2)

b) Determine its numerical value:

Length = (1.602 x 10^-19 C)^2/(9.109 x 10^-31 kg * (3.00 x 10^8 m/s)^2)

Length = 5.29 x 10^-11 m

Question:

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n1​). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Answer:

1. Frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n1) can be calculated using the Rydberg formula:

Frequency = R * (1/n1^2 - 1/n^2)

Where R is the Rydberg constant.

2. For large n, the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n1) can be approximated as:

Frequency = R * (1/n^2)

3. This frequency equals the classical frequency of revolution of the electron in the orbit which can be calculated using the equation:

Frequency = (2 * π * e^2)/(m * n^2 * h)

Where e is the charge of the electron, m is the mass of the electron, n is the principal quantum number, and h is Planck’s constant.

4. By equating the two equations, we can obtain an expression for the classical frequency of revolution of the electron in the orbit:

R = (2 * π * e^2)/(m * n^2 * h)

Thus, the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n1) for large n equals the classical frequency of revolution of the electron in the orbit.