11. How to apply a perpendicular magnetic field to a quantum ring?
12.Is it possible that some of the "Spooky" things we observe at the Quantum level could be explained by a possible Time Bias?
13.Which is the 'epsilon' of our local universe?
fast...............................................
Answers
Answer:
Actually I work on the semiconductor quantum rings, theoretically. There are lots of theoretical and experimental probes in which, people have applied a magnetic filed perpendicular to quantum rings and studied Aharonove-Bohm effect on such systems. But as I have never been in a lab, I can't understand how people are technically able to apply such a uniform magnetic field on a quantum ring with about 100nm diameter.
Explanation:
Given that , In an Airthmetic Progression [ A.P. ] sum of it's n th terms is 3n²+5n & it's k th terms is 164 .
Exigency To Find : The value of k ?
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
\begin{gathered}\maltese\:\:\: \sf Let's \:say \:that \:, \:\pmb{\bf S_n } \: be \:the \:sum \:of \;n \:terms \:of \:an \:A.P\: & \:,\:\\ \sf \pmb{\sf a_n } \:be \: n^{th} \:term \:of \:an \:A.P \: \:[ \:Airthmetic \:Progression \:] \:\\\\\end{gathered}✠Let′ssaythat,SnSnbethesumofntermsofanA.PananbenthtermofanA.P[AirthmeticProgression],
Given that ,
Sum of n terms of an A.P is 3n²+5n .
\begin{gathered}\qquad \sf \therefore \;\: S_n \: = \: 3n^2 \: + \: 5n \: \\\\\qquad \dashrightarrow \:\sf \;\: S_n \: = \: 3n^2 \: + \: 5n \: \\\\\end{gathered}∴Sn=3n2+5n⇢Sn=3n2+5n
Therefore,
Sum of n - 1 terms of an Airthmetic Progression [ A.P. ] .
\begin{gathered}\qquad \sf \therefore \;\: S_{ n- 1 } \: = \: 3( n - 1 )^2 \: + \: 5( n - 1 ) \: \\\\\qquad \dashrightarrow \sf \;\: S_{ n- 1 } \: = \: 3( n - 1 )^2 \: + \: 5( n - 1 ) \: \\\\\qquad \dashrightarrow \sf \;\: S_{ n- 1 } \: = \: 3n^2 - \: n - 2 \: \\\\\end{gathered}∴Sn−1=3(n−1)2+5(n−1)⇢Sn−1=3(n−1)2+5(n−1)⇢Sn−1=3n2−n−2
Now ,
As , We know that ,
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ a_n \: =\: S_n - \: S_{n-1} }\bigg\rgroup \\\\\qquad \dashrightarrow \sf a_n \: =\: S_n - \: S_{n-1} \:\:\\\\\end{gathered}†⎩⎪⎪⎪⎧an=Sn−Sn−1⎭⎪⎪⎪⎫⇢an=Sn−Sn−1
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}⋆NowBySubstitutingtheknownValues:
\begin{gathered}\qquad \dashrightarrow \sf a_n \: =\: S_n - \: S_{n-1} \:\:\\\\\qquad \dashrightarrow \sf a_n \: =\: \Big\{ 3n^2 + 5n \:\Big\} - \: \Big\{ 3n^2 - n - 2 \Big\} \:\:\\\\\qquad \dashrightarrow \sf a_n \: =\: 3n^2 + 5n \: - \: 3n^2 + n + 2 \:\:\\\\\qquad \dashrightarrow \sf a_n \: =\: 6n + 2 \:\:\\\\\dashrightarrow \underline {\boxed{\pmb{\frak{\purple { a_n \: =\: 6n + 2 }}}}}\:\\\\\end{gathered}⇢an=Sn−Sn−1⇢an={3n2+5n}−{3n2−n−2}⇢an=3n2+5n−3n2+n+2⇢an=6n+2⇢an=6n+2an=6n+2
Therefore,
\sf k^{th}kth term will be 6k + 2 .
AND ,
In an Airthmetic Progression [ A.P. ] \sf k^{th}kth term is 164 .
\begin{gathered}\qquad \qquad \sf \leadsto \;\: a_k \: = \: 6k +2 \: \& ,\:\\\\\qquad \sf \leadsto \;\: a_k \: = \: 164 \: \\\\\qquad \sf \therefore \;\: 164 \: = \: 6k +2 \: \\\\\qquad \sf \dashrightarrow \;\: 164 \: = \: 6k +2 \: \\\\\qquad \sf \dashrightarrow \;\: 164 - 2 \: = \: 6k \: \\\\\qquad \sf \dashrightarrow \;\: 162 \: = \: 6k \: \\\\\qquad \sf \dashrightarrow \;\: k \: = \: \dfrac{162 }{6} \: \\\\\qquad \sf \dashrightarrow \;\: k \: = \: 27 \: \\\\\dashrightarrow \underline {\boxed{\pmb{\frak{\purple { k \: =\: 27 }}}}}\:\\\\\end{gathered}⇝ak=6k+2&,⇝ak=164∴164=6k+2⇢164=6k+2⇢164−2=6k⇢162=6k⇢k=6162⇢k=27⇢