Bohr S Model Of Hydrogen Atom

The total energy E of an electron in a hydrogen atom is the sum of its kinetic energy K and its potential energy U. According to the Bohr model of the hydrogen atom, the total energy of the electron is given by:

E = K + U

The kinetic energy K is equal in magnitude to the potential energy, but opposite in sign. Therefore, we have:

K = -U

The total energy of the electron is:

E = K + (-K) = -K

Thus, the ratio of the total energy E to the kinetic energy K is:

E / K = -K / K = -1

Correct Answer:

Correct Answer: (A) -1

Recall the Bohr Model Energy Levels for Hydrogen:
According to the Bohr model, the energy of an electron in the orbit of a hydrogen atom is given by the formula: where is the principal quantum number ( ).

Determine the Principal Quantum Number (n):
We are given that the total energy of the electron is eV. We can use the energy formula to find the value of corresponding to this energy level: Divide both sides by -1 eV: Solve for : Take the square root (since must be positive): So, the electron is in the second energy level (n=2).

Recall Bohr's Angular Momentum Quantization Condition:
One of the postulates of the Bohr model is that the angular momentum of an electron orbiting the nucleus is quantized and is given by: where is the principal quantum number and is Planck's constant.

Calculate the Angular Momentum:
Substitute the value into the angular momentum formula: Simplify the expression:

The angular momentum of the electron in the state with energy -3.4 eV (which corresponds to n=2) is .

So, the correct option (A):

Given:

• Balmer series formula for hydrogen spectrum:
• We need to find and for the longest wavelength in the Balmer series.

Step 1: Understand the Balmer Series

The Balmer series corresponds to transitions where the electron falls to the level. Thus, is fixed as 2.

Step 2: Determine for Longest Wavelength

The longest wavelength occurs when the energy difference is smallest, i.e., for the smallest possible transition:

Substituting and :

This gives the smallest , and thus the largest .

Conclusion:

The values of and for the longest wavelength in the Balmer series are and .

Answer: