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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) (m_(e) =9.11 × 10^(-31) kg). |
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Answer» Solution :WAVELENGTH of photon of X-ray `lambda=1Å=10^(-10)m` `h=6.63xx10^(-34)JS,c=3xx10^(8)ms^(-1)` Energy of one photon, `E=hv=(hc)/(lambda)` `therefore E=(6.63xx10^(-34)xx3xx10^(8))/(10^(-10))` `therefore E=19.89xx10^(-16)J` `therefore E=(19.89xx10^(-16))/(1.6xx10^(-19))eV` `=12.43xx10^(3)eV~~12.4KeV` ........(1) `implies` For electron `lambda=1Å=10^(-10)m` `m=9.1xx10^(-31) kg` `therefore "momentum p"=(h)/(lambda)=(6.63x10^(-34))/(10^(-10))` `therefore p=6.63xx10^(-24)kg ms^(-1)` `implies` Kinetic energy of eleectron, `K=(p^(2))/(2m)` `therefore K=((6.63xx10^(-24))^(2))/(2xx9.1xx10^(-31))` `therefore K=2.4152xx10^(-17)J` `therefore K=(2.4152xx10^(-17))/(1.6xx10^(-19))` `therefore K=1.5-95xx10^(2)` `therefore K=150.9 eV` From equation (1) and (2) For equal (same) wavelength energy of photon is very ,arge compared to energy of electron. |
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