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| 1. |
An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron. |
| Answer» <html><body><p></p>Solution :Here `lamda=1.00nm=1.00xx10^(-9)m` <br/> (a) Momentum of electron momentum of photon `p=(h)/(lamda)=(6.63xx10^(-34))/(1.00xx10^(-9))=6.63xx10^(-25)<a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a>" "<a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>^(-1)`. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) Energy of photon `E=(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/(lamda)p*c=6.63xx10^(-25)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>)=1.989xx10^(-16)J` <br/> `=(1.989xx10^(-16))/(1.6xx10^(-19))eV=1.24xx10^(3)eV or 1.24keV`. <br/> (c) Kinetic energy of electron `K=(p^(2))/(2m)=((6.63xx10^(-25))^(2))/(2xx9.11xx10^(-31))=2.413xx10^(-19)J` <br/> `=(2.413xx10^(-19))/(1.6xx10^(-19))eV=1.51eV`.</body></html> | |