Respuesta :
Answer:
KE = 160.41 keV
Explanation:
given data
radii = 1.00 cm
radii = 2.40 cm
uniform magnetic field of magnitude = 0.0540 T
to find out
energy (in keV) of the incident electron
solution
first we find the velocity of charge with radii 1 cm then 2.40 cm by given formula
R = [tex]\frac{mv}{qB}[/tex] ......................1
put here value
1 × [tex]10^{-2}[/tex] = [tex]\frac{9.1*10^{-31}*v}{1.6*10^{-19}*0.052}[/tex]
solve we get
v = 91428571 m/s
and with radii 2.40 from equation 1
R = [tex]\frac{mv}{qB}[/tex]
2.4 × [tex]10^{-2}[/tex] = [tex]\frac{9.1*10^{-31}*v1}{1.6*10^{-19}*0.052}[/tex]
solve we get
v1 = 219428570 m/s
so
now we calculate energy of incident electron
KE = 0.5 m (v² + v1² )
KE = 0.5 ×9.1×[tex]10^{-31}[/tex]× (91428571² + 219428570² )
KE = 2.57 × [tex]10^{-14}[/tex] J
KE = 2.57 × [tex]10^{-14}[/tex] J × 6.242 × [tex]10^{15}[/tex]
KE = 160.41 keV
