The magnetic field $(B)$ due to revolving electron at the center is proportional to $\frac{Z^{3}}{n^{x}}$ . The value of $x$ is -

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Solution

The correct option is C $5$
The magnetic field $(B)$ due to revolving electron at the center,

$B \propto \frac{Z^{3}}{n^{5}}$

$∴ x = 5$

Hence, option $(C)$ is the correct answer.

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Similar questions

(

n

Z

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Column I	Column II
$(A)$ Due to revolving electron, the magnetic field produced at its centre is proportional to	$(p) n^{- 5}$
$(B)$ Magnetic moment of a revolving electron is proportional to	$(q) n$
$(C)$ De-Broglie wavelength of a revolving electron is proportional to	$(r) Z^{3}$
$(D)$ Areal velocity of a revolving electron about the nucleus is proportional to	$(s)$ Independent of $Z$
	$(t) Z^{- 1}$

= Z

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n^{t h}

Magnetic ﬁeld at the centre (at nucleus) of the hydrogen like atoms (atomic number = Z) due to the motion of electron in nth orbit is proportional to

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