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Find the ratio of longest wavelangth in ...

Find the ratio of longest wavelangth in Paschen series to shortest wavelength in Balmer series.

Text Solution

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We have,
Case I : `R=12Omegal_(2)=120cm`
Case II : `R=18Omega,l_(2)=150cm`
From first condition,
Internal resistance (r) `=R((l_(1)-l_(2))/(l_(2)))`
`r=12((l_(1)-l_(2))/(120))" "......(i)`
From second condition,
Internal resistance (r) `=R((l_(1)-l_(2))/(l_(2)))`
`=18((l_(1)-150)/(150))" "......(ii)`
From equations (i) and (ii)
`12((l_(1)-120)/(120))=18((l_(1)-150)/(150))`
`5(l_(1)-120)=6(l_(1)-150)`
`5l_(1)-600=6l_(1)-900`
`l_(1)=300cm.
Putting the value of `l_(2)` in equation (i) we, get
`r=12((300-120)/(120))`
`r=18Omega`
Therefore, the balancing lenght is 300 cm and internal resistance is `18Omega`.
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