Question

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (mₑ=9.1 1 X 10⁻³¹ kg). ​

Answer 1

Solution:-

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•For x-photon of wavelength of 1Å:

$\bullet \sf \lambda = 1 Å=10 ^{ - 10} m$

We know that ,

$\sf E= \dfrac{hc}{ \lambda}$

$\sf \: \: \: \: = \dfrac{6.62 \times 10 ^{ - 34} \times 3 \times 10^{8} }{10^{ - 10} }$

$\sf \: \: \: \: = 1.986 \times 10^{ - 15} j$

$\sf \: \: \: \: = \dfrac{1.986 \times {10}^{ - 15} }{1.6 \times {10}^{ - 19} }$

$\: \: \: \: \sf = 12.4 \times 10^{3} eV$

$\: \: \: \: \sf = 12.4Kev$

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•For the electron of wavelength 1Å:

$\sf \bullet\lambda = 1Å=10^{-10}m$

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Now, we know that,

$\sf \lambda = \dfrac{h}{mv}$

$\sf \: or \: mv = \dfrac{h}{ \lambda}$

$\: \: \: \: \sf = \dfrac{6.62 \times {10}^{ - 34} }{ {10}^{ - 10} }$

$\: \: \: \: \sf = 6.62 \times 10^{ - 24} kg \: ms ^{ - 1}$

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•Energy of the electron:

$\sf E= \dfrac{1}{2} m {v}^{2}$

$\: \: \: \: \sf = \dfrac{ \: \: (mv) ^{2} \: \: }{ \: \: 2m \: \: \: }$

$\: \: \: \: \sf = \dfrac{ \: (6.62 \times {10}^{ - 24} )^{2} \: \: \: }{2 \times 9.11 \times 10^{ - 31 \: \: } }$

$\: \: \: \: \sf = 2.405 \times {10}^{ - 17} j$

$\: \: \: \: \sf = \dfrac{2.405 \times {10}^{ - 17} }{1.6 \times 10^{ - 19} }$

$\: \: \: \: \bf= 150.3eV$

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It follows that x-ray photon has greater energy

Answer 2

Kindly Refer to Attachment !!!