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11.Dual Nature of Radiation and matter
medium
In a Thomson set-up for the determination of e/m, electrons accelerated by $2.5$ $kV$ enter the region of crossed electric and magnetic fields of strengths $3.6 \times {10^4}V{m^{ - 1}}$ and $1.2 \times {10^{ - 3}}T$ respectively and go through undeflected. The measured value of $e/m$ of the electron is equal to
A
$1.0 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
B
$1.76 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
C
$1.80 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
D
$1.85 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
Solution
(c) $\frac{e}{m} = \frac{{{E^2}}}{{2V{B^2}}} = \frac{{{{(3.6 \times {{10}^4})}^2}}}{{2 \times 2.5 \times {{10}^3} \times {{(1.2 \times {{10}^{ – 3}})}^2}}}$
$ = 1.8 \times {10^{11}}C/kg$.
Standard 12
Physics
