Question

In a hydrogen atom following the Bohr's postulates the product of linear momentum and angular momentum is proportional to \((\mathrm{n})^{x}\) where ' \(n^{\prime}\) is the orbit number. Then ' \(x^{\prime}\) is : (1) 0 (2) 2 (3) \(-2\) (4) 1

Short answer

0

Step-by-step solution

- Understand Bohr's Postulates

Bohr's postulates state that the electron revolves around the nucleus in certain stable orbits without radiating energy. The angular momentum of the electron in these orbits is quantized and is given by: \[L = n\frac{h}{2\text{π}}\] where \(L\) is the angular momentum, \(n\) is the orbit number, and \(h\) is Planck's constant.

- Linear Momentum Calculation

The linear momentum \(p\) of an electron in an orbit of radius \(r\) is related to its angular momentum \(L\) by the equation: \[L = p \times r\]. Thus, we have: \[p = \frac{L}{r}\].

- Determine Relationship Between Radius and Orbit Number

In a hydrogen atom, the radius of the \(n\)-th orbit is given by: \[r_n = n^2 \times r_1\] where \(r_1\) is the radius of the first orbit.

- Substitute and Simplify

Using the expressions for \(L\) and \(r_n\), substitute \(r_n\) in the linear momentum expression: \[p = \frac{n\frac{h}{2\text{π}}}{n^2 \times r_1} = \frac{h}{2\text{π} \times n \times r_1}\].

- Combine Linear and Angular Momentum

Now, calculate the product of linear momentum \(p\) and angular momentum \(L\): \[p \times L = \frac{h}{2\text{π} \times n \times r_1} \times n\frac{h}{2\text{π}} = \frac{h^2}{4\text{π}^2 \times r_1} = \text{constant}\].

- Conclude the Value of x

The result shows that the product of linear momentum and angular momentum is a constant, independent of the orbit number \(n\). Therefore, the value of \(x\) is 0.

Key concepts

Angular Momentum Quantization

Bohr's postulates introduced the idea that the angular momentum of an electron in a hydrogen atom is not continuous but quantized. This means electrons can only occupy certain orbits where their angular momentum, denoted as \(\text{L}\), is given by the formula: \(\text{L} = n\frac{h}{2\text{π}}\). Here, \(n\) is the orbit number, and \(h\) is Planck's constant.
The implication of this quantization is that an electron cannot exist in arbitrary energy levels but only at specific levels determined by the quantized angular momentum. This leads to the stability of atoms because electrons revolve in stable orbits without emitting radiation.

Linear Momentum

The linear momentum \(\text{p}\) of an electron in an orbit is related to its angular momentum \(\text{L}\) and the radius of its orbit \(\text{r}\). The relationship is expressed as: \(L = p \times r\). By rearranging this, we find the linear momentum as: \(p = \frac{L}{r}\).
Therefore, the linear momentum is the angular momentum divided by the radius of the orbit. This provides a direct connection between the movement of an electron along its path (linear momentum) and its circular motion around the nucleus (angular momentum).
This relationship helps in better understanding the forces and velocities involved in the electron's orbit.

Orbit Radius Relationship

In Bohr's model, the radius \(\text{r}_n\) of the \(n\)-th orbit of an electron in a hydrogen atom is given by: \(r_n = n^2 \times r_1\), where \(r_1\) is the radius of the first orbit. This shows that the radius increases with the square of the orbit number.
Thus, as electrons move to higher energy levels (orbits with larger \(n\)), the radius of these orbits increases significantly.
When combined with linear momentum, this helps to show why the product of linear momentum and angular momentum results in a value that is constant and does not depend on \(n\).
Therefore, it further proves that the value of \(x\) in the given problem is 0, indicating the independence of this product from the orbit number.