An electron of mass 9-1times 10-31-kg and charge 1-6times 10-19-C is accelerated through a potential difference of -V- volt- The de Broglie wavelength -lambda - associated with the electron isdfrac-12-27-V-A-0-12-27sqrt-V-A-0-dfrac-12-27-sqrt-V-A-0-dfrac-1-12-27sqrt-V-A-0-

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Toppr Admin · Accepted Answer

12mv2-eVmv-x221A-2meVNow-xA0- -x3BB-hmv-xA0- -xA0- -xA0- -xA0- -h-x221A-2meV-xA0- -xA0- -xA0- -xA0- -xA0-12-27-x221A-VA0

An electron of mass 9.1×10−31kg and charge 1.6×10−19C is accelerated through a potential difference of 'V' volt. The de Broglie wavelength (λ) associated with the electron is12.27VA012.27√VA012.27√VA0112.27√VA0

12mv2=eVmv=√2meVNow λ=hmv =h√2meV =12.27√VA0

$\frac{1}{2} m v^{2} = e V$
$m v = \sqrt{2 m e V}$
Now $λ = \frac{h}{m v}$
$= \frac{h}{\sqrt{2 m e V}}$
$= \frac{12.27}{\sqrt{V}} A^{0}$