the inverse of ratio of number of revolutions of electron per second in third and fourth of H-atom in terms of velocity (v) and radius (r) is

To find the inverse of the ratio of the number of revolutions of an electron per second in the third and fourth energy levels of a hydrogen atom, we need to start by understanding the relationship between the electron's velocity, radius, and the number of revolutions it makes in a given time frame. This involves some fundamental concepts from atomic physics and circular motion.

Understanding Electron Motion in Hydrogen Atom

In a hydrogen atom, electrons occupy different energy levels, which can be described by quantum numbers. The third energy level corresponds to \( n = 3 \) and the fourth to \( n = 4 \). The velocity of an electron in a given orbit can be derived from the Bohr model of the hydrogen atom, which states that the velocity \( v \) of an electron in the nth orbit is given by:

v = Z \cdot \frac{e^2}{2 \epsilon_0 h} \cdot \frac{1}{n}

Here, \( Z \) is the atomic number (which is 1 for hydrogen), \( e \) is the elementary charge, \( \epsilon_0 \) is the permittivity of free space, and \( h \) is Planck's constant. The radius \( r \) of the nth orbit is given by:

r = n^2 \cdot \frac{h^2}{4 \pi^2 m e^2}

where \( m \) is the mass of the electron. For hydrogen, we can simplify these equations to find the velocity and radius for the third and fourth orbits.

Calculating Velocity and Radius

• For \( n = 3 \):
• Velocity: \( v_3 = \frac{Z \cdot e^2}{2 \epsilon_0 h} \cdot \frac{1}{3} \)
• Radius: \( r_3 = 3^2 \cdot \frac{h^2}{4 \pi^2 m e^2} = 9 \cdot r_1 \) (where \( r_1 \) is the radius of the first orbit)
• For \( n = 4 \):
• Velocity: \( v_4 = \frac{Z \cdot e^2}{2 \epsilon_0 h} \cdot \frac{1}{4} \)
• Radius: \( r_4 = 4^2 \cdot \frac{h^2}{4 \pi^2 m e^2} = 16 \cdot r_1 \)

Finding the Number of Revolutions

The number of revolutions per second (frequency) for an electron in a circular orbit can be expressed as:

f = \frac{v}{2 \pi r}

Now, we can calculate the frequencies for both orbits:

• For \( n = 3 \):

Substituting \( v_3 \) and \( r_3 \):

f_3 = \frac{v_3}{2 \pi r_3} = \frac{\frac{Z \cdot e^2}{2 \epsilon_0 h} \cdot \frac{1}{3}}{2 \pi (9 \cdot r_1)} = \frac{Z \cdot e^2}{54 \pi \epsilon_0 h r_1}

• For \( n = 4 \):

Substituting \( v_4 \) and \( r_4 \):

f_4 = \frac{v_4}{2 \pi r_4} = \frac{\frac{Z \cdot e^2}{2 \epsilon_0 h} \cdot \frac{1}{4}}{2 \pi (16 \cdot r_1)} = \frac{Z \cdot e^2}{128 \pi \epsilon_0 h r_1}

Calculating the Ratio and Its Inverse

Now, we can find the ratio of the frequencies:

\(\frac{f_3}{f_4} = \frac{\frac{Z \cdot e^2}{54 \pi \epsilon_0 h r_1}}{\frac{Z \cdot e^2}{128 \pi \epsilon_0 h r_1}} = \frac{128}{54} = \frac{64}{27}\)

The inverse of this ratio is:

\(\frac{f_4}{f_3} = \frac{27}{64}\)

Final Expression in Terms of Velocity and Radius

Thus, the inverse of the ratio of the number of revolutions of the electron per second in the third and fourth energy levels of the hydrogen atom can be expressed as:

\(\frac{f_4}{f_3} = \frac{27}{64}\)

This means that for every 64 revolutions the electron makes in the third orbit, it makes 27 revolutions in the fourth orbit. This relationship highlights how the energy levels affect the motion of electrons in an atom.