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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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Calculate wavelength of emitted characteristic X-ray from a tungsten (Z=74) target when an electron drops from an M shell to a vacancy in K shell

$(a)\;1.8 \times 10^{-12} m\qquad(b)\;1.8 \times 10^{-13} m\qquad(c)\;1.8 \times 10^{-11} m\qquad(d)\;1.8 \times 10^{-9} m$

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Answer : (c ) $\;1.8 \times 10^{-11} m$
Explanation :
Tungston is multi electron atom. Due to shielding of nuclear charge by -ve charge of inner core electrons each electron is subjected to an effective nuclear charge Z H which is different for different shells.
Thus energy of an electron in $\;n^{th}\;$ level of a multi electron atom is given by
$\varepsilon_{n}=-\large\frac{13.6 Z^2 H}{n^2} eV$
For an electron in K shell (n=1) , ZH=(Z-1)
Thus energy of electron in K- shell is :
$\varepsilon_{K}=-\large\frac{(74-1)^2 \times13.6}{1^2}=-72500eV$
For an electron in M shell (n=3) the nucleus is shielded by one electron of n=1 state and eight electrons of n=2 state , a total of nine electrons ,So that ZH=Z-9 . Thus energy of an electron in M shell is :
$\varepsilon_{M}=-\large\frac{(74-9)^2\times13.6}{3^2}=-6380eV$
The emitted X-ray photon has an energy given by
$h \nu=\varepsilon_{M}-\varepsilon_{K}=-6380 eV-(-72500eV)$
$=66100eV$
or $\;\large\frac{hc}{\lambda}=66100 \times 1.6 \times10^{-19} J$
$\lambda=\large\frac{hc}{66100\times1.6\times10^{-19}} m$
$=\large\frac{(6.63\times 10^{-34})\times(3 \times 10^{8})}{66100\times1.6\times10^{-19}}\;m$
$=0.0188\times10^{-9} m=1.8\times10^{-11} m\;.$
answered Feb 26, 2014 by yamini.v
 

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