physics - An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure ( 110^{-2} mm of Hg). A magnetic field of 2.83 *10^{-4} T. curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure ( 110^{-2} mm of Hg). A magnetic field of 2.83 *10^{-4} T. curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data

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Anonym0us (Student, UG1 Grade)

Qalaxia Knowlege Bot (last edited 7 months ago)

I found an answer from large.stanford.edu

NASA SP-413

Nov 17, 1975 ... The concept of spacecraft Earth, a sphere of finite resources and ominous ... for ten weeks to construct a convincing picture of how ... Appendix K — The Lunar Gas Gun Mass Driver . ... that space colonization offers large potential benefits and ... kPa or ~100 mm Hg) for good respiration yet low enough to.

For more information, see NASA SP-413

Qalaxia Master Bot (last edited 7 months ago)

I found an answer from en.wikipedia.org

Glossary of physics - Wikipedia

This glossary of physics is a list of definitions of terms and concepts relevant to physics, its sub-disciplines, and related fields, including mechanics, materials ...

For more information, see Glossary of physics - Wikipedia

Pravalika (last edited 7 months ago) (Student, UG2 Grade)

Given that

Potential of gun collector = 100 V

Magnetic fiels B = 2.83 * 10^{-4} T

Radius of the curved path r = 12.0 cm = 12 *10^2 m

Step 1: Determine the velocity of an ejected electron from gun due to voltage

Kinetic energy of electron K.E = \frac{1}{2} mv^2

v^2 = \frac{2 K.E}{m}

According to energy conversion K.E = energy of each electron = eV

v^2 = \frac{2 eV}{m} ....................(1)

Step 2: Velocity of electron in a spherical bulb due to magnetic field.

Electron moving in a circular path so it acquires centripetal force

F = \frac{mv^2}{r}

Force on the electron in a magnetic field

F =evB \because \theta = 90\degree

So, from the two equations we can write, evB = \frac{mv^2}{r}

eB = \frac{mv}{r}

v = \frac{eBr}{m}

v^2 = \frac{e^2B^2r^2}{m^2} ....................(2)

Step 3: Calculating the specific charge ratio

From equation (1) and (2)

\frac{e^2B^2r^2}{m} = \frac{2 eV}{m}

\frac{e}{m} = \frac{2V}{B^2 r^2}

Plug in the values in the above equation.

[math] \frac{e}{m} = \frac{2 * 100}{[2.83 * 10^{-4}]^2 (12*10^2)^2} [/math]

\frac{e}{m} = 1.734 * 10^{11} C/kg

Hence, the specific charge ratio \frac{e}{m} = 1.734 * 10^{11} C/kg