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An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?
de -Broglie wavelength,`lambda = h/(sqrt(2meV))`
Where,
`m ->\text { Mass of electron} = 9.1 xx 10^-31kg`
`lambda = (6.626 xx 10^-34)/(sqrt (2 xx 9.1 xx 10^-31 xx 1.6 xx 10^-19 xx 64))`
`lambda = (6.626 xx 10^-34)/(sqrt 1863.63, xx 10^-50)`
`lambda = (6.626 xx 10^-34)/(sqrt 1063.68 xx 10^-25)`
`=6.626/43.17 xx 10^-9`
`=0.15 xx 10^-9 m`
` =1.5 xx 10^-10 m`
This value of wavelength corresponds to the X-ray region of the electromagnetic spectrum.
Concept: de-Broglie Relation
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Text Solution
Solution : Here, <br> `therefore` de-Broglie wavelength is `V=64` volts <br> `lambda =(1.227)/(sqrt(V))nm =(1.227)/(sqrt(64)) =(1.227)/(8)=0.1533 nm`. <br> This wavelength is associated with X-rays (range :1 nm `10 ^-3` nm)
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