Question

An electron moves in a circular orbit of radius 0.51 Å around a nucleus at a frequency of 6.8 x 10^15 Hz: Estimate the magnetic field at the nucleus.​

Answer 1

Answer:

easy one .................

Answer 2

The magnetic field at the nucleus is $2.43\times 10^5\ T$.

Explanation:

Given that,

Radius of the circular orbit, $r=0.51\ A=0.51\times 10^{-10}\ m$

Frequency of the electron, $f=6.8\times 10^{15}\ Hz$

To find,

Th magnetic field at the nucleus.

Solution,

The frequency with which an electron moves in the magnetic field is given by :

$f=\dfrac{qB}{2\pi m}$

m and q are the mass and charge on proton

$B=\dfrac{2\pi mf}{q}$

$B=\dfrac{2\pi \times 9.1\times 10^{-31}\times 6.8\times 10^{15}}{1.6\times 10^{-19}}$

$B=2.43\times 10^5\ T$

So, the magnetic field at the nucleus is $2.43\times 10^5\ T$. Hence, this is the required solution.

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