Question

An electron moves from rest position from an electron and reaches with velocity of 10^7m/s to another electronic in cathode ray tube calculate p.d between electrodes(me=1.9×10^-31)??? .
I will mark as branleist​

Answer 1

Answer:

Explanation:

       use energy conservation, 1/2 mv² = Vq

                    v is velocity , V is p.d and q = 1.6 x 10 to power -19 (charge)

             solve and u will get the answer as 118.75 volts

              hope u get it n let me know..........

  PLZ MARK AS BRAINLIEST...........

Answer 2

• Charge of electron, $\sf{Q=1.6\times10^{-19}\ C}$

• Mass of electron, $\sf{m=9.1\times10^{-31}\ kg}$

• Velocity of electron, $\sf{v=10^7\ m\,s^{-1}}$

The energy acquired by the electron is given by,

$\longrightarrow\sf{E=QV}$

where $\sf{V}$ is the potential difference between the electrodes which has to be found out.

The energy acquired by the electron is stored in it as its kinetic energy since it's in motion. Thus,

$\longrightarrow\sf{\dfrac{1}{2}\,mv^2=QV}$

$\longrightarrow\sf{V=\dfrac{mv^2}{2Q}}$

$\longrightarrow\sf{V=\dfrac{9.1\times10^{-31}\left(10^7\right)^2}{2\times1.6\times10^{-19}}}$

$\longrightarrow\sf{\underline{\underline{V=284.375\ V}}}$

Hence the potential difference is $\bf{284.375\ V.}$