Question

The ratio of angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of hydrogen atom is ...... \(\\{\mathrm{A}\\} \sqrt{2}\) \(\\{B\\} \sqrt{(1 / 2)}\) \(\\{\mathrm{C}\\}(1 / 2) \quad\\{\mathrm{D}\\} 2\)

Short answer

The ratio of the angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of the hydrogen atom is \(\frac{1}{2}\). Therefore, the correct answer is (C) \(\frac{1}{2}\).

Step-by-step solution

Write down the formula for angular momentum in an allowed orbit

The formula for the angular momentum (L) in the allowed orbits according to Bohr's quantization condition is given by: \( L = n\frac{h}{2\pi} \) Where \(n\) is the orbit number (integer), and \(h\) is Planck's constant.

Calculate the angular momentum for the first allowed orbit

Let's find the angular momentum (L1) for the first allowed orbit (n=1). \( L_1 = 1\frac{h}{2\pi} = \frac{h}{2\pi} \)

Calculate the angular momentum for the second allowed orbit

Now, let's find the angular momentum (L2) for the second allowed orbit (n=2). \( L_2 = 2\frac{h}{2\pi} = \frac{2h}{2\pi} \)

Find the ratio of the angular momenta

Let's find the ratio of L1 to L2. \( \textup{Ratio} (\frac{L_1}{L_2}) = \frac{\frac{h}{2\pi}}{\frac{2h}{2\pi}} \)

Simplify the ratio

Simplify the ratio by cancelling out the common terms. \( \frac{L_1}{L_2} = \frac{\frac{h}{2\pi}}{\frac{2h}{2\pi}} = \frac{1}{2} \)

Select the correct answer from the options given

The correct answer is the one that matches our calculated ratio. The ratio of the angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of the hydrogen atom is \(\frac{1}{2}\). So, the correct answer is: (C) \(\frac{1}{2}\).