Question
Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.
✓
Answer
Dipole moment $(\mu)$$=\text{niA}=\frac{1\times\text{q}}{\text{tA}}=\text{qfA}$
$=\text{e}\times\frac{\text{me}^4}{4\in_0^2\text{h}^3\text{n}^3}\times\big(\pi\text{r}^2_0\text{n}^2\big)=\frac{\text{me}^5\times\big(\pi\text{r}^2_0\text{n}^2\big)}{4\in_0^2\text{h}^3\text{n}^3}$
$=\frac{\big(9.1\times10^{-31}\big)\big(1.6\times10^{-19}\big)^5\times\pi\times(0.53)^2\times10^{-20}\times1}{4\times\big(8.85\times10^{-12}\big)^2\big(6.64\times10^{-34}\big)^3(1)^3}$
$=0.0009176\times10^{-20}=9176\times10^{-24}\text{Am}^2$
$=\text{e}\times\frac{\text{me}^4}{4\in_0^2\text{h}^3\text{n}^3}\times\big(\pi\text{r}^2_0\text{n}^2\big)=\frac{\text{me}^5\times\big(\pi\text{r}^2_0\text{n}^2\big)}{4\in_0^2\text{h}^3\text{n}^3}$
$=\frac{\big(9.1\times10^{-31}\big)\big(1.6\times10^{-19}\big)^5\times\pi\times(0.53)^2\times10^{-20}\times1}{4\times\big(8.85\times10^{-12}\big)^2\big(6.64\times10^{-34}\big)^3(1)^3}$
$=0.0009176\times10^{-20}=9176\times10^{-24}\text{Am}^2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The weight of an empty boat is 500 kg. When it floats 1/3 part of its sink in water. Find out what is the maximum weight can be loaded on the boat?
The specification on a heater coil is $250V, 500W.$ Calculate the resistance of the coil. What will be the resistance of a coil of $1000W$ to operate at the same voltage?
→Define kinetic energy density of a wave. Derive an expression for its maximum value. Using it, prove that the intensity of the wave is, $\text{I}=\frac{1}{2}\rho \omega^2\text{A}^2\nu.$
Ans.
→Ans.
For a particle in S.H.M., the displacement x of the particle as a function of time t is given as $\text{x = A}\sin(2\pi\text{t}).$ Here x is in cm and t is in seconds.
Let the time taken by the particle to travel from x = 0 to $\text{x}=\frac{\text{A}}{2}$ be $T_1$ and the time taken to travel from $\text{x}=\frac{\text{A}}{2}$ to x = A be $T_2$. Find $\frac{\text{T}_1}{\text{T}_2}.$
→Let the time taken by the particle to travel from x = 0 to $\text{x}=\frac{\text{A}}{2}$ be $T_1$ and the time taken to travel from $\text{x}=\frac{\text{A}}{2}$ to x = A be $T_2$. Find $\frac{\text{T}_1}{\text{T}_2}.$
Prove that the torque acting due to a force F in the xy plane is, $\tau_{\text{z}}=\text{x}\text{F}_{\text{y}}-\text{y}\text{F}_{\text{x}'}$
→A current of 5.0A exists in the circuit shown in the figure. The wire PQ has a length of 50cm and the magnetic field in which it is immersed has a magnitude of 0.20T. Find the magnetic force acting on the wire PQ.
A simple microscope has a magnifying power of $3.0$ when the image is formed at the near point $(25\ cm)$ of a normal eye.
→- What is its focal length?
- What will be its magnifying power if the image is formed at infinity?
The following equation represents standing wave set up in medium,
$\text{y}=4\cos\frac{\pi\text{x}}{5}\sin40\pi\text{t},$
where x and y are in cm and t in sec. Find out the amplitude and the velocity of the two component waves and calculate the distance between adjacent nodes. What is the velocity of a medium particle at x = 3cm at time $\frac{1}{8}\text{sec}$?
→$\text{y}=4\cos\frac{\pi\text{x}}{5}\sin40\pi\text{t},$
where x and y are in cm and t in sec. Find out the amplitude and the velocity of the two component waves and calculate the distance between adjacent nodes. What is the velocity of a medium particle at x = 3cm at time $\frac{1}{8}\text{sec}$?
Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?
An ideal gas expands from $100\ cm^3$ to $200\ cm^3$ at a constant pressure of $2.0 \times 10^5 Pa$ when $50J$ of heat is supplied to it. Calculate,
→