Saved Bookmarks
| 1. |
A nerrow beamof singly charged potassium ions of kinetic energy 32keVis injected into a region of width 1.00 cm having a magnetic field of strenght 0.500 T as shown in . The ions arecollected at a screen 95.5cm away from the field region. Ifthe beam contains isotopes of atomic weights 39 and 41, find the separaton between the points where these isotopes strike the screen. Take the mass of a potassium ion= A (1.6 X 10^(-27) kg where A is the mass number. |
|
Answer» Solution :For E-39 ` m = 39 xx 1.6 xx 10^(-7) ) kg , B= 5 xx 10^(-1) T ,` ` q = 1.6 xx 10^(-10)_(c) , KE = 32 kev` Velocity projection ` (1)/(2) xx 3 sin thetaxx (1.6 xx 10^(-27) )` , rArr`V^(2)= 32 xx 10^(2) xx 106 xx 10^(-19) ` ` V= 4.05 xx 10^(6) ` Through out the motion the horiantal velocity remation constant i = 24 xx 10^(-18) sec ` [ time taken to crossthe magnetic field ] Accle in theregion having magneticfield . `= 5193 , 53xx 10 ^(4) m/s ^(2) ` V (in vertical direction ) = at ` 5193.53 xx 10^(8) xx 24 xx 10^(-18) ` ` = 12464.48 m/s ` Totaltime taken to reach the screen ` = 0.000002382 sec`, Time gap= 2382 xx 10^(-0) - 24 xx 10^(-2) sec` , ` = 2358 xx 10^(-2) sec` Distance moved VERTICALLY (in the time) ` = 12464 .48 xx 2358 xx 10^(-0) ` = 0.0293m ` ` v^(2) = 2aS` Net display fromline ` = 0.0001495 + 0.0293912` ` = 0.0295407 on` for K-41l : (1)/(2)xx 41 xx 1.6 xx 10^(-17) v^(2)` ` = 32 xx 10^(0) xx 1.6 xx 10^(-0)` lt.brgt ` rArr v = 39 . 509 m/s ` ` a = 4818.193 xx 10^(0) m/s ^(2)` i = ( time taken for COMMING outside from magnetic field ltbbbrgt `= 25 xx 10(-0) sec` V = at (verticalvelocity ) ` = 4818. 193 xx 10^(8) xx 25 xx 10^(-0)` `= 12045.48 m/s `ltlbrgt ( time TOTAL to reach the screen ) `= 0.000002442 ` `time gap = 2442 xx 10^(-0) -25 xx 10^(-8)` `= 2417 xx 10^(-8)` distance moved vericaily `= 12045.48 xx 2417 xx 10^(-0)` ` = 0.0291` Now `V^(2) = 2aS = (2045.48)^(2)` `2 xx 4818.19 xx 10^(6)S` `rArr S = 0.0001505 m` Net distance traveiled `= 0.0001505 + 0.0291139` `= 0.0292644` Net gap between `K- 39 and K-41` `= 0.0295407 - 0.0292644` `0.0001763 m = 0.27 mm` . |
|