A narrow beam of singly charged potassium ions of kinetic energy 32 ke V is injected into a region of width 1.00 cm having a magnetic field of strength 0.500 T as shown in figure (34.E16). The ions are collected at a screen 95.5 cm 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 (1.6×10−27) kg where A is the mass number.
A narrow beam of singly charged potassium ions of kinetic energy 32 ke V is injected into a region of width 1.00 cm having a magnetic field of strength 0.500 T as shown in figure (34.E16). The ions are collected at a screen 95.5 cm 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 (1.6×10−27) kg where A is the mass number.
For K-39 :
m=39×1.6×10−27 kg,
B=5×10−1 T,
q=1.6×10−19 c, KE=32 kev
Velocity projection :
12×39×(1.6×10−27)
⇒ V2=32×103×1.6×10−19
V=4.05×105
Through out the motion the horizontal velocity remains constant
t=24×10−19 sec
[time taken to cross the magnetic field]
Accln in the region having magnetic field
=5193.53×108 m/s2
V (in vertical direction) = at
=5193.53×108×24×10−19
=12464.48 m/s
Total time taken to reach the screen
=0.000002382 sec.
Time gap = 2382×10−9−24×10−9 sec.
=2358×10−9 sec.
Distance moved vertically (in the time)
=12464.48×2358×10−9
=0.0293 m
v2=2aS
⇒ (12464.48)2=2×5193.53×108×S
⇒ 0.1495×10−3=S
Net display from line
=0.0001495+0.0293912
=0.0295407 m
For K-41l :
12×41×1.6×10−27 v2
=32×103×1.6×10−9
⇒ v=39.509 m/s
a=4818.193×108 m/s2
t = (time taken for comming outside from magnetic field)
=25×10−9 sec.
V = at (vertical velocity)
=4818.193×108×25×10−9
=12045.48 m/s
(time total to reach the screen)
=0.000002442
time gap = 2442×10−9−25×10−9
=2417×10−9
distance moved vertically
=12045.48×2417×10−9
=0.0291
Now, V2=2aS=(2045.48)2
=2×4818.19×108 S
⇒ S=0.0001505 m
Net distance travelled
=0.0001505+0.0291139
=0.0292644
Net gap between K - 39 & K - 41
=0.0295407−0.0292644
=0.0001763 m=0.27 mm
For K-39 :
m=39×1.6×10−27 kg,
B=5×10−1 T,
q=1.6×10−19 c, KE=32 kev
Velocity projection :
12×39×(1.6×10−27)
⇒ V2=32×103×1.6×10−19
V=4.05×105
Through out the motion the horizontal velocity remains constant
t=24×10−19 sec
[time taken to cross the magnetic field]
Accln in the region having magnetic field
=5193.53×108 m/s2
V (in vertical direction) = at
=5193.53×108×24×10−19
=12464.48 m/s
Total time taken to reach the screen
=0.000002382 sec.
Time gap = 2382×10−9−24×10−9 sec.
=2358×10−9 sec.
Distance moved vertically (in the time)
=12464.48×2358×10−9
=0.0293 m
v2=2aS
⇒ (12464.48)2=2×5193.53×108×S
⇒ 0.1495×10−3=S
Net display from line
=0.0001495+0.0293912
=0.0295407 m
For K-41l :
12×41×1.6×10−27 v2
=32×103×1.6×10−9
⇒ v=39.509 m/s
a=4818.193×108 m/s2
t = (time taken for comming outside from magnetic field)
=25×10−9 sec.
V = at (vertical velocity)
=4818.193×108×25×10−9
=12045.48 m/s
(time total to reach the screen)
=0.000002442
time gap = 2442×10−9−25×10−9
=2417×10−9
distance moved vertically
=12045.48×2417×10−9
=0.0291
Now, V2=2aS=(2045.48)2
=2×4818.19×108 S
⇒ S=0.0001505 m
Net distance travelled
=0.0001505+0.0291139
=0.0292644
Net gap between K - 39 & K - 41
=0.0295407−0.0292644
=0.0001763 m=0.27 mm
