Riya Andrews

2022-10-09

Calculate the de Broglie wavelength of electrons accelerated through 3868 V. Round off the answer to 2 decimal places with scientific representation.

Krha77

Beginner2022-10-10Added 8 answers

Given data: -

The potential difference is V = 3868 V.

Here, the mass of the electron is $m=9.1\times {10}^{-31}kg$, the value of the Planck's constant is $h=6.62\times {10}^{-34}$ J-s and the charge on electron is $e=1.6\times {10}^{-19}$ C.

The expression for the DE Broglie wavelength of electrons is given as,

$\lambda =\frac{h}{\sqrt{2meV}}$

Substitute all the known values in above equation,

$\lambda =\frac{6.62\times {10}^{-34}J\cdot s}{\sqrt{2\times 9.1\times {10}^{-31}kg\times 1.6\times {10}^{-19}C\times 3868V}}\phantom{\rule{0ex}{0ex}}\lambda =1.97\times {10}^{-11}m$

Thus, the DE Broglie wavelength of electrons is $\lambda =1.97\times {10}^{-11}m$

The potential difference is V = 3868 V.

Here, the mass of the electron is $m=9.1\times {10}^{-31}kg$, the value of the Planck's constant is $h=6.62\times {10}^{-34}$ J-s and the charge on electron is $e=1.6\times {10}^{-19}$ C.

The expression for the DE Broglie wavelength of electrons is given as,

$\lambda =\frac{h}{\sqrt{2meV}}$

Substitute all the known values in above equation,

$\lambda =\frac{6.62\times {10}^{-34}J\cdot s}{\sqrt{2\times 9.1\times {10}^{-31}kg\times 1.6\times {10}^{-19}C\times 3868V}}\phantom{\rule{0ex}{0ex}}\lambda =1.97\times {10}^{-11}m$

Thus, the DE Broglie wavelength of electrons is $\lambda =1.97\times {10}^{-11}m$

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