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Questions  >>  CBSE XII  >>  Physics  >>  Electromagnetic waves
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Q)

State the parts of e.m spectrum to which $14.4\;kev$ (energy of a particular transition $Cn Fe^{57} $ nucleus associated with a famous high resolution spectroscopic method Massbauer spectroscopy ]

$\begin{array}{1 1} micro\;waves \\ radio\;waves \\ u.v\;rays \\ x-rays \end{array} $

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A)
Solution :
Here Energy $E= [4.4 Kev = 14.4 \times 10^3 \times 1.6 \times 10^{-19}]$
$E= h \alpha$
$\alpha= \large\frac{E}{h}$
$\qquad= \large\frac{14.4 \times 10^3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\qquad= 3 \times 10^{11} \;MHZ$
Hence this frequency lies in the x-ray region of the e.m spectrum.
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