JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 27)
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)
$${n^{ - 4}}$$
$${n^{ - 5}}$$
n$$-$$3
n$$-$$2
Explanation
Magnetic field at the center of neucleus of H-atom
B = $${{{\mu _0}I} \over {2{r_n}}}$$
Radius of nth orbital,
rn = $${{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}$$
$$\therefore\,\,\,$$ rn $$ \propto $$ n2
velocity of electron in nth orbital,
$$\upsilon $$n = $$\left( {{{{e^2}h} \over {2{\varepsilon _0}}}} \right){Z \over n}$$
$$\therefore\,\,\,$$ $$\upsilon $$n $$ \propto $$ n$$-$$1
I = $${q \over t}$$ = $${e \over {{{2\pi {r_n}} \over {{\upsilon _n}}}}}$$ = $${{e{\upsilon _n}} \over {2\pi {r_n}}}$$
$$\therefore\,\,\,$$ B = $${{{\mu _0}.\left( {{{e{\upsilon _n}} \over {2\pi {r_n}}}} \right)} \over {2{r_n}}}$$ = $${{{\mu _0}\,e{\upsilon _n}} \over {4\pi r_n^2}}$$
$$\therefore\,\,\,$$ B $$ \propto $$ $${{{\upsilon _n}} \over {r_n^2}}$$
$$ \Rightarrow $$ $$\,\,\,$$ B $$ \propto $$ $${{{n^{ - 1}}} \over {{n^4}}}$$
$$ \Rightarrow $$$$\,\,\,$$ B $$ \propto $$ n $$-$$5
B = $${{{\mu _0}I} \over {2{r_n}}}$$
Radius of nth orbital,
rn = $${{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}$$
$$\therefore\,\,\,$$ rn $$ \propto $$ n2
velocity of electron in nth orbital,
$$\upsilon $$n = $$\left( {{{{e^2}h} \over {2{\varepsilon _0}}}} \right){Z \over n}$$
$$\therefore\,\,\,$$ $$\upsilon $$n $$ \propto $$ n$$-$$1
I = $${q \over t}$$ = $${e \over {{{2\pi {r_n}} \over {{\upsilon _n}}}}}$$ = $${{e{\upsilon _n}} \over {2\pi {r_n}}}$$
$$\therefore\,\,\,$$ B = $${{{\mu _0}.\left( {{{e{\upsilon _n}} \over {2\pi {r_n}}}} \right)} \over {2{r_n}}}$$ = $${{{\mu _0}\,e{\upsilon _n}} \over {4\pi r_n^2}}$$
$$\therefore\,\,\,$$ B $$ \propto $$ $${{{\upsilon _n}} \over {r_n^2}}$$
$$ \Rightarrow $$ $$\,\,\,$$ B $$ \propto $$ $${{{n^{ - 1}}} \over {{n^4}}}$$
$$ \Rightarrow $$$$\,\,\,$$ B $$ \propto $$ n $$-$$5
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