Answer:
Complete question
In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.
a. What electric field strength, in volts per mater, is needed to select a speed of 4.2 x 10^6 m/s?
b. What is the voltage, in kilovolts, between the plates if they are separated by 0.95 cm?
Explanation:
Given that,
magnetic field B = 0.095T
Speed of particle v = 4.2 ×10^6m/s
Separation between plate d = 0.95cm
d = 0.95/100 = 0.0095m
a. Using the mass spectrometer velocity selector relationship between the electric field and magnetic field.
v = E/B
Where
v is the speed selector
B is magnetic field
E is electric field
Therefore, E = vB
E = 4.2 × 10^6 × 0.095
E = 0.399× 10^6
E = 3.99 × 10^5 V/m
b. Voltage?
The relationship between electric field and potential difference between the two plates is given as
V = Ed
V = 3.99 × 10^5 × 0.0095
V = 3790.5 V
To kV, 1kV = 1000V
Then, V = 3.7905kV
V ≈ 3.791 kV
