According to Bohr's theory, the angular momentum of an electron in 5th orbit is

According to Bohr's Theory of Hydrogen Atoms-like atoms, the angular momentum L of an electron in the nth orbit is quantized and given by:

\[ L = mvr = \frac{n \times h}{2π} \]

Where:

• L = Angular momentum of the electron
• m = Mass of the electron
• v = Velocity of the electron
• r = Radius of the orbit
• n = Principal quantum number, which denotes the number of the orbit (1 for the first orbit, 2 for the second, and so on)
• h = Planck's constant

Using this rule, the angular momentum of an electron in the nth orbit is:

\[ L_n = \frac{n \times h}{2π} \]

Now, let's calculate the angular momentum for the electron in the 5th orbit, i.e., n=5:

\[ L_5 = \frac{5 \times h}{2π} \]

To simplify:

\[ L_5 = 2.5 \frac{h}{π} \]

Hence, the angular momentum of an electron in the 5th Bohr orbit is \( 2.5 \frac{h}{π} \).

The Correct Option is D \( 2.5 \frac{h}{π} \).

More Question on Bohr's Theory:

Bohr's Theory & Angular Momentum in Hydrogen Atom Expressions for Kinetic Energy in Bohr's Theory Factors Bohr's Theory Can't Explain Limitations of Bohr's Theory Understanding Bohr's Model for Hydrogen Atom Limitations of Bohr's Model Explained Calculating Electron Energy with Bohr's Theory