According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to : (n=principal quantum number)

  • Option 1)

     n−4

  • Option 2)

    n−5

  • Option 3)

    n−3

  • Option 4)

    n−2

 

Answers (1)

As we learnt in

Velocity of electron in nth orbital -

v= \left ( \frac{e^{2}}{2\epsilon_{0}h} \right )\frac{z}{n}
 

- wherein

v\: \alpha \frac{z}{n}

\frac{e^{2}}{2\epsilon_{0}h}= \frac{c}{137}

 

 

Radius of nth orbital -

r_{n}= \frac{\epsilon _{0}n^{2}h^{2}}{\pi mZe^{2}}

 

- wherein

r_{n}\alpha \: \frac{n^{2}}{Z}

\frac{\epsilon_{0}h^{2}}{\pi me^{2}}= 0.529A^{\circ}

 

 Magnetic field due to current at centre of circle =\frac{\mu_{0}I}{2\pi r}

I=\frac{q}{t}=\frac{e}{\left(\frac{2\pi r}{v} \right )}=\frac{ev}{2\pi r}

B=\frac{\mu.\left(\frac{ev}{2\pi r} \right )}{2r}=\frac{\mu_{0}ev}{4\pi r^{2}}

B=\frac{\left(\frac{\mu_{0}e}{4\pi} \right ).\left(\frac{c}{137} \right )\frac{z}{n}}{\left(r_{0}\frac{n^{2}}{z} \right )^{2}}=\frac{\mu_{0}ec}{4\pi\times137r_{0}^{2}}\times\frac{z^{3}}{n^{5}}

B\alpha\ n^{-5}

Correct answer is 2


Option 1)

 n−4

This is an incorrect option.

Option 2)

n−5

This is the correct option.

Option 3)

n−3

This is an incorrect option.

Option 4)

n−2

This is an incorrect option.

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