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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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Where $\;0.5 A^{0}\;$ x-rays strike a material , the photoelectrons from K shell are observed to move in a circle of radius 23mm in a magnetic fields of $\;2\times10^{-2}T\;.$ Than the binding energy of K-shell electrons is

$(a)\;1.3 KeV\qquad(b)\;4.2 KeV\qquad(c)\;5.8 KeV\qquad(d)\;6.2 KeV$

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Answer : (d) $\;6.2 KeV$
Explanation :
The velocity of photo electrons is found
$F=ma$
$eVB=\large\frac{mV^2}{R} \quad \; V=\large\frac{e}{m}\;BR$
The kinetic energy of photo electrons is then
$K=\large\frac{1}{2}\;mV^2=\large\frac{1}{2}\;\large\frac{e^2B^2R^2}{m}$
$=\large\frac{1}{2}\;\large\frac{(1.6 \times 10^{-19}C)^2\;(2 \times10^{-2}T)^2\;(23 \times10^{-3} m)^2}{(9.1\times10^{-31} kg)}$
$=2.97\times10^{-15} J$
$or\; K=(2.97\times10^{-15} J)\;\large\frac{(1 Kev)}{(1.6\times10^{-16}) J}=18.36eV$
The energy of incident photon is
$\varepsilon \nu =\large\frac{hc}{\lambda}=\large\frac{12.4 KeV A^{0}}{0.5A^{0}}=24.8 eV$
The binding energy is difference between these two valves
$BE=\varepsilon \nu-K=24.8-18.6=6.2 KeV\;.$
answered Feb 23, 2014 by yamini.v
 

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