Kinetic energy of an electron in the second Bohr orbit | Chemca Solution

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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 _________. (Nearest integer)

(Where \( a_0 \) is the Bohr radius and \( m \) is the mass of the electron).

Given: \( \pi = 3.14 \)

Enter the value of \( 10x \) (integer only):

Detailed Step-by-Step Solution

To solve this, we must derive the kinetic energy of an electron in a Bohr orbit in terms of Planck's constant (\(h\)), mass (\(m\)), and the Bohr radius (\(a_0\)).

Step 1: Relate Velocity to the Bohr Postulates

From Bohr's quantization of angular momentum, we know:

\( mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr} \)

For a Hydrogen atom, the radius of the \(n^{\text{th}}\) orbit is given by:

\( r_n = n^2 a_0 \)

Step 2: Find the Velocity in the 2nd Orbit

Substitute \( n = 2 \) into our equations:

\( r_2 = (2)^2 a_0 = 4a_0 \)

Now, substitute \( r_2 \) and \( n=2 \) back into the velocity equation:

\( v_2 = \frac{2h}{2\pi m (4a_0)} = \frac{h}{4\pi m a_0} \)

Step 3: Calculate Kinetic Energy

The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). Let's plug in our velocity \( v_2 \):

\( KE = \frac{1}{2} m \left( \frac{h}{4\pi m a_0} \right)^2 \)

\( KE = \frac{1}{2} m \left( \frac{h^2}{16\pi^2 m^2 a_0^2} \right) \)

\( KE = \frac{h^2}{32\pi^2 m a_0^2} \)

Step 4: Find \( x \) and Calculate \( 10x \)

The problem states that \( KE = \frac{h^2}{x m a_0^2} \). Comparing this to our derived expression:

\( \frac{h^2}{x m a_0^2} = \frac{h^2}{32\pi^2 m a_0^2} \)

Therefore, \( x = 32\pi^2 \).

Now, we need to calculate \( 10x \) using the given value of \( \pi = 3.14 \):

\( 10x = 10 \times (32\pi^2) = 320 \times (3.14)^2 \)

\( 10x = 320 \times 9.8596 \)

\( 10x = 3155.072 \)

Conclusion: The calculated value is 3155.072. Rounding to the nearest integer, the final answer is 3155.

Mastering Bohr Model Derivations

In JEE Main, questions often require you to derive standard formulas (like Kinetic Energy or Total Energy) in terms of fundamental constants rather than just plugging in numbers. Memorizing that \( v \propto \frac{Z}{n} \) and \( r \propto \frac{n^2}{Z} \) is a great starting point for these derivations.

To practice more algebraic derivations involving energy, velocity, and radius in the Bohr Model, be sure to read our comprehensive guide on the Structure of Atom Class 11 Chemistry.

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