Question

qn below:

Answer 1

\[\large v_n=\dfrac{e}{\sqrt{4 \pi \in mr_n}}\]

\[\large m v_n r_n=\dfrac{nh}{2 \pi}\]

I need to get expression for v_n and r_n combining these equations .

Answer 2

divide the second by the first. you should get an equation in r_n only

Answer 3

@sirm3d ...y not solve in "draw" and help him....

Answer 4

why don't you draw it, @Koikkara ? feel free to join us.

Answer 5

but the answer i have doesnt match

\[r_n=(\dfrac{n^2}{m})(\dfrac{h}{2 \pi }^2 )\dfrac{4 \pi \in }{e^2}\]

Answer 6

\[v_n=\dfrac{1}{n} \dfrac{e^2}{4 \pi \in} \dfrac{1}{(h/2 \pi)}\]

Answer 7

\[\Large{\frac{m^2v_n^2r_n^2}{v_n^2}=\frac{n^2h^24\pi\varepsilon m r_n}{4\pi^2e^2}\\r_n=\frac{n^2h^2\varepsilon}{e^2m}=\frac{n^2}{m}\frac{h^2}{\pi}\frac{\varepsilon }{e^2}}\]

Answer 8

if you multiply the middle fraction by 4pi/4pi, \[\Large \frac{h^2}{\pi}=\frac{h^24\pi}{4\pi^2}=\left(\frac{h}{2\pi}\right)^24\pi\]

Answer 9

oh i seee..then what about v_n ?

Answer 10

in the second equation
\[\Large v_n=\frac{nh}{2\pi m}\cdot \frac{1}{r_n}\]

Answer 11

then how i bring the other terms ?

Answer 12

should i substitute for r_n ?

Answer 13

\[\Large v_n=\frac{nh}{2\pi m}\frac{m}{n^2}\left(\frac{2\pi}{h}\right)^2\frac{e^2}{4\pi\varepsilon}=\frac{e^2}{2nh\varepsilon }\]

Answer 14

ok thanks a lot ..i spent a lot of time on this ....

Answer 15

you can rewrite the denominator as \[\Large 2nh\varepsilon =n(4\pi)(\frac{h}{2\pi})\varepsilon\]

Answer 16

ok

Answer 17

this symbol \(\varepsilon\) is \varepsilon in TeX

Answer 18

ok ... does this method of combining equations like these have a name ?

Answer 19

i am coming across this type of solving for first time

Answer 20

just manipulation of symbols.

Answer 21

if algebra, if a=b and c=d, then a/c = b/d provided none of the denominators is zero.

Answer 22

ok thanks !