Question

Electrons are accelerated in the picture tube of a television through potential difference of 8.00 * 10 ^ 3 V. (Use the values q e =1.60*10^ -19 kg and m e =9.109*10^ -31 kg.)
Please please helpppp?

Answer 1

Answer:

$\lambda=1.37\times 10^{-11}\ m$

Explanation:

Give that,

The potential difference of the electrons, $V=8\times 10^{3}\ V$

We need to find the wavelength of the electrons.

Using the conservation of energy,

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

Put all the values,

$\lambda=\sqrt{\dfrac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 8\times 10^3}}\\\\\lambda=1.37\times 10^{-11}\ m$

So, the wavelength of the electrons is $1.37\times 10^{-11}\ m$.