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InterviewSolution
| 1. |
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)=9.11xx10^(-31)kg). |
| Answer» <html><body><p></p>Solution :It is given here that wavelength of probe `lamda=1Å=1xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)m` <br/> When we use X-rays as the probe, the energy of X-ray photon <br/> `E_("X-ray")=hv=(hc)/(lamda)=(6.63xx10^(-34)xx3xx10^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>))/(1xx10^(-10))J=(6.63xx10^(-34)xx3xx10^(8))/(1xx10^(-10)xx1.6xx10^(-19))eV=12.4xx10^(3)eV=12.4keV` <br/> When <a href="https://interviewquestions.tuteehub.com/tag/electron-968715" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRON">ELECTRON</a> waves are used as the probe, the energy of an electron <br/> `E_("electron")=(1)/(2)mv^(2)=(p^(2))/(2m)=(1)/(2m)(h^(2))/(lamda^(2))=((6.63xx10^(-34))^(2))/(2xx.911xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/31-306830" style="font-weight:bold;" target="_blank" title="Click to know more about 31">31</a>)xx(1xx10^(-10))^(2))`J <br/> `=((6.63xx10^(-34))^(2))/(2xx9.11xx10^(-31)xx(10^(-10))^(2)xx1.6xx10^(-19))eV=150eV` <br/> A <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> comparison of two answers shows `E_("X-ray")gtgtE_("electron")`.</body></html> | |