A narrow beam of singly charged potassium ions of kinetic energy 32keV is injected into a region of width 1.00 cm having a magnetic field of strength 0.500 T as shown in . The ions are collected at a screen 95.5cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation 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.

Explanation

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 theta xx (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 cross the magnetic field ]
Accle in the region having magnetic field .
` = 5193 , 53 xx 10 ^(4) m/s ^(2) `
V (in vertical direction ) = at
` 5193.53 xx 10^(8) xx 24 xx 10^(-18) `
` = 12464.48 m/s `
Total time 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 from line
` = 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 (vertical velocity )
` = 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` .