Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to \(\frac{h^2}{x m a_0^2}\). The value of \(10x\) is _________. (\(a_0\) is radius of Bohr's orbit)
(Nearest integer)
[ Given : \(\pi = 3.14\) ]

Hint:

Substitute \(v = \frac{nh}{2\pi mr}\) into \(KE = \frac{1}{2}mv^2\) to get \(KE = \frac{n^2 h^2}{8 \pi^2 m r^2}\). Then substitute \(r = n^2 a_0\) to simplify the expression.

Answer 1

Correct Answer: 3155

Step 1: Understanding the Concept:
The kinetic energy (\(KE\)) of an electron in a Bohr orbit is given by \(\frac{1}{2}mv^2\). We need to express this in terms of Planck's constant \(h\), mass \(m\), and Bohr radius \(a_0\).
Step 2: Key Formula or Approach:
1. Bohr's quantization of angular momentum: \(mvr = \frac{nh}{2\pi}\)
2. Radius of \(n^{th}\) orbit for Hydrogen: \(r_n = n^2 a_0\)
Step 3: Detailed Explanation:
For the second orbit (\(n = 2\)), the radius is:
\[ r_2 = 2^2 a_0 = 4 a_0 \]
From the angular momentum quantization:
\[ mv = \frac{nh}{2\pi r} = \frac{2h}{2\pi (4 a_0)} = \frac{h}{4\pi a_0} \]
The kinetic energy is:
\[ KE = \frac{1}{2} m v^2 = \frac{(mv)^2}{2m} = \frac{1}{2m} \left( \frac{h}{4\pi a_0} \right)^2 \]
\[ KE = \frac{h^2}{2m \cdot 16 \pi^2 a_0^2} = \frac{h^2}{32 \pi^2 m a_0^2} \]
Comparing this with the given formula \(\frac{h^2}{x m a_0^2}\):
\[ x = 32 \pi^2 \]
Given \(\pi = 3.14\):
\[ x = 32 \times (3.14)^2 = 32 \times 9.8596 = 315.5072 \]
We need the value of \(10x\):
\[ 10x = 10 \times 315.5072 = 3155.072 \]
Step 4: Final Answer:
The nearest integer value for \(10x\) is 3155.