1.

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

Answer»

`35xx10^(-12)m`
35 Å
3.5 NM
3.5 Å

Solution :`35xx10^(-12)m`
`E=hf=(hc)/(LAMBDA)`
`THEREFORE lambda=(hc)/(E )`
`=(6.625xx10^(-34)xx3xx10^(8))/(35xx10^(3)xx1.6xx10^(-19)) `
`=0.3549xx10^(-10)m`
`=35.49xx10^(-12)m`


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