Question

If the potential difference used to accelerate electrons is tripled by what factor does the de Broglie wavelength of electron beam change?

Answer 1

New wavelength will become old wave length by √3....coz wavelength = h/√2√m√q√v and this is because

Answer 2

Given that,

The potential difference used to accelerate electrons is tripled.

To find,

The change in De-Broglie wavelength.

Solution,

The relation between the De-Broglie wavelength and potential difference used to accelerate electrons is given by :

$\lambda\propto \dfrac{1}{\sqrt{V} }$

V is potential difference

So,

$\dfrac{\lambda_1}{\lambda_2}=\sqrt{\dfrac{V_2}{V_1}}$

Here, $V_2=3V_1$

i.e.

$\dfrac{\lambda_1}{\lambda_2}=\sqrt{\dfrac{3V_1}{V_1}}\\\\\dfrac{\lambda_1}{\lambda_2}=\sqrt3\\\\\lambda_2=\dfrac{\lambda_1}{\sqrt{3} }$

So, the new wavelength reduced by a factor of $\dfrac{1}{\sqrt{3} }$.

Learn more,

De-Broglie wavelength

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