What is the ratio of wavelength of last line of Balmer series and last line of Lyman series?

36.

A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π2 Ω, the total charge flowing through the coil during this time is

C.

32 μC

Text Solution

`2``1``4``0.5`

Answer : C

Solution : Using the relation <br> `1/lambda=R(1/(n_(1)^(2))-1/(n_(2)^(2)))` <br> For last line of Blamer series, `n_(1) = 2, n_(2) = oo` <br> `(1)/(lambda_(B)) = R((1)/(2^(2))-(1)/(oo))=R/4.....(i)` <br> for last line of Balmer series, `n_(1)=2, n_(2)=oo` <br> `1/(lambda_(L))=R(1/(1^(2))-1/(oo))=R.........(ii)` <br> Dividing (ii) by (i), we get <br> `(lambda_(B))/(lambda_(L))=4`