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प्रश्न
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 (∼10−2 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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उत्तर
Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B = 2.83 × 10−4 T
Radius of the circular orbit r = 12.0 cm = 12.0 × 10−2 m
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e.,
`1/2 "mv"^2 = "eV"`
`"v"^2 = (2"eV")/"m"` ..............(1)
It is the magnetic field, due to its bending nature, that provides the centripetal force
`("F" = "mv"^2/"r")` for the beam. Hence, we can write:
Centripetal force = Magnetic force
`"mv"^2/"r" = "evB"`
`"eB" = "mv"/"r"`
`"v" = ("eBr")/"m"` ..............(2)
Putting the value of v in equation (1), we get:
`(2"eV")/"m" = ("e"^2"B"^2"r"^2)/"m"^2`
`"e"/"m" = "2V"/("B"^2"r"^2)`
= `(2 xx 100)/((2.83 xx 10^(-4))^2 xx (12 xx 10^(-2))^2)`
= 1.73 × 1011 C kg−1
Therefore, the specific charge ratio (e/m) is 1.73 × 1011 C kg−1.
संबंधित प्रश्न
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