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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 22 of 37
Marks: +1, -0
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 (10−210^{-2} mm of Hg). A magnetic field of 2.83 × 10−410^{-4} T curves the path of the electrons in a circular orbit of radius 12.0 cm. Determine e/m from the data.
Solution:  
Here V = 100 V, r = 12 cm = 12 × 10−210^{-2} m, B = 2.83 × 10−410^{-4} T
The gain in kinetic energy of an electron when accelerated through V volts is
12mv2\frac{1}{2} m v^{2} = eV or v2v^{2} = 2eVm\frac{2eV}{m}
As the magnetic field provides centripetal force to the electron, therefore,
evB = mv2r\frac{mv^{2}}{r} or v = eBrm\frac{eBr}{m}
v2v^{2} = e2B2r2m2\frac{e^{2}B^{2}r^{2}}{m^{2}}
∴ 2eVm\frac{2eV}{m} = e2B2r2m2\frac{e^{2}B^{2}r^{2}}{m^{2}}
Specific Charge , em\frac{e}{m} = 2VB2r2\frac{2V}{B^{2}r^{2}} = 2×100(2.83×10−4)2×(12×10−12)2\frac{2 \times 100}{(2.83 \times 10^{-4})^{2} \times (12 \times 10^{-12})^{2}}
= 1.73 × 1011 CKg−110^{11}\ \mathrm{CKg}^{-1}
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