1.

Wavelength of photon having 35 keV energy will be……..(h=6.625xx10^(-34))J-s,c=3xx10^(8)ms^(-1),1eV=1.6xx10^(-19)J).

Answer»

`35xx10^(-12)m`
35Å
3.5nm
3.5Å

Solution :`E=hf=(hc)/(LAMBDA)`
`therefore =(hc)/(E )=(6.625xx10^(-34)xx3xx10^(8))/(35xx10^(3)xx1.6xx10^(-19))`


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