The answer is -7 of, but how? Figure 27.3 o 0.20 T deutron source! Va accelerato

Question

The answer is -7 of, but how?

Figure 27.3 o 0.20 T deutron source! Va accelerator grid A mass spectrograph is operated with deuterons, which have a charge of +e and a mass of 3.34 x 10-27 kg. Deuterons emerge from the source, which is grounded with negligible velocity. The velocity of the deuterons as they pass through the accelerator grid is 8.0 x 105 m/s. A uniform magnetic field of magnitude B 0.20 T, directed out of the plane, is present at the right of the grid. 5) In Figure 273, the electric potential of the accelerator grid Va is closest to: D) +9 kv 5) A) +7 kV B) -9 kV C)-7kV E) +11 kV

Explanation / Answer

Solution:

For the deutrons to pass through the accelerator with a straight path:

1/2mv^2 = eV

=> V = 1/2(3.34*10^-27)(8*10^5)2 /(1.6*10-19)

=> V = potential difference = 6680 (approx.)

i.e around 6.68 7 kV

(negative as the charge is positive!)

=> -7kV

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