Question 166: Atoms & Nuclei

Question

According to Bohr's theory, the moment of momentum of an electron revolving in $$4^{	ext {th }}$$ orbit of hydrogen atom is:

Accepted Answer

According to Bohr's theory, one of the postulates specifies that the angular momentum of an electron in orbit around a nucleus is quantized. This quantization can be expressed by the formula:
$$ L = n\frac{h}{2\pi} $$
Where:
$L$ is the angular momentum of the electron,
$n$ is the principal quantum number (or the orbit number in simpler terms), which can take positive integer values (1, 2, 3, ...),
$h$ is Planck's constant ($6.62607015 	imes 10^{-34} m^2 kg / s$), and
$\frac{h}{2\pi}$ is often denoted as $\hbar$ (h-bar), known as the reduced Planck's constant.
For an electron in the 4th orbit ($n = 4$) of a hydrogen atom, we substitute $n = 4$ into the equation:
$$ L = 4\frac{h}{2\pi} $$
Therefore, the moment of momentum (or angular momentum) of an electron in the $4^{	ext{th}}$ orbit of a hydrogen atom is:
$$ L = 4\frac{h}{2\pi} = 2\frac{2h}{2\pi} = 2\frac{h}{\pi} $$
Hence, the correct option is:
Option A: $2 \frac{h}{\pi}$

Answer

$$2 \frac{h}{\pi}$$

Answer

$$\frac{h}{2 \pi}$$

Answer

$$\frac{h}{\pi}$$

Answer

$$8 \frac{h}{\pi}$$

Atoms & Nuclei

According to Bohr's theory, the moment of momentum of an electron revolving in $4^{\text {th }}$ orbit of hydrogen atom is:

Choose the correct answer: