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

`\lambda=1.37* 10^(-11)\ m`

Step-by-step explanation:

Give that,

The potential difference of the electrons,
`V=8* 10^(3)\ V`

We need to find the wavelength of the electrons.

Using the conservation of energy,

`2meV=(h^2)/(\lambda^2)\\\\\lambda=\sqrt{(h^2)/(2meV)}`

Put all the values,

`\lambda=\sqrt{((6.63* 10^(-34))^2)/(2* 9.1* 10^(-31)* 1.6* 10^(-19)* 8* 10^3)}\\\\\lambda=1.37* 10^(-11)\ m`

So, the wavelength of the electrons is
`1.37* 10^(-11)\ m`.