In a Thomson's e/m experiment , with an accelerating potential of 150V and a deflecting electric field of magnitude 6*10^6 N/C , the electrons travel in a straight line .

i)At what fraction of the speed of light do the electrons move?
ii)what magnitude of magnetic field do you need?

In a double-slit interference experiment, the wavelength is 500 nm, the slit separation is 0.150 mm, and the screen is 33.7 cm away from the slits. What is the linear distance between adjacent maxima on the screen?

To determine the fraction of the speed of light at which the electrons move in a Thomson's e/m experiment, we can use the formula:

v = (2eV / m)^(1/2)

Where:
- v is the speed of the electrons,
- e is the charge of an electron (approximated as 1.6 x 10^-19 C),
- V is the accelerating potential (150V in this case),
- m is the mass of an electron (approximated as 9.11 x 10^-31 kg).

Let's plug in the values and solve the equation:

v = (2 * (1.6 x 10^-19 C) * (150V) / (9.11 x 10^-31 kg))^(1/2)

v ≈ 6.13 x 10^6 m/s

The speed of light (c) is approximately 3 x 10^8 m/s. To find the fraction of the speed of light, we can divide the speed of the electrons (v) by the speed of light (c):

Fraction of the speed of light = v / c

Fraction of the speed of light ≈ (6.13 x 10^6 m/s) / (3 x 10^8 m/s)
≈ 0.0204 or 2.04%

Therefore, the electrons are moving at approximately 2.04% of the speed of light.

To determine the magnitude of the magnetic field (B) required, we can use the formula:

B = (eV) / (Bd)

Where:
- B is the magnitude of the magnetic field,
- d is the distance between the plates, and
- B is the magnetic field strength that balances the electric field (6 x 10^6 N/C in this case).

Let's rearrange the formula to solve for B:

B = (eV) / (Bd)

Substituting the given values:

B = ((1.6 x 10^-19 C) * (150V)) / ((6 x 10^6 N/C) * d)

Unfortunately, the distance between the plates (d) is not provided in the question. To find the magnitude of the magnetic field, we need this information.