Question

An electron is accelerated through a potential difference of 64 volt what is the de broglie wavelength associated with it to which part of the electromagnetic spectrum does this value of wavelength correspond

Answer 1

de broglie wavelength :------

$λ = \frac{h}{ \sqrt[]{2mev} }$


$= \frac{6.82 \times 10 { }^{ - 34} }{ \sqrt{2 \times 1.6 \times 10 {}^{ - 19} \times v\times 9.1 \times 10 {}^{ - 31 } } }$


$= \: \frac{12.27}{ \sqrt{v} } \times 10 {}^{ - 10} m \:$

$here \: v = 64v$


$λ = \frac{12.27}{ \sqrt{64} } Å \: = \frac{12.27}{8} Å \: = 1.53 \: Å \:$


THIS CORRESPONDS TO X-RAY REGION OF THE ELECTROMAGNETIC SPECTRUM

Answer 2

The De Broglie wavelength of the electrons is 1.53 Angstrom.

Explanation:

Given that,

An electron is accelerated through a potential difference of 64 volt.

To find,

The De Broglie wavelength

Solution,

The De broglie wavelength is given by :

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

On solving,

$\lambda=\dfrac{12.27}{\sqrt{V} }A$

Here, V = 64 volts

$\lambda=\dfrac{12.27}{\sqrt{64} }A$

$\lambda=1.53\ A$

So, the De Broglie wavelength of the electrons is 1.53 Angstrom. The wavelength of x rays varies from 10 to 0.1 Angstrom. So, it correspond to x rays.

Learn more,

Electromagnetic spectrum

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