Question

An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?

Answer 1

Answer:

See below.

Explanation:

According to the question, we know that,

work done is given by,  $W=qV$

and change in kinetic energy is, Δ $KE=W=1=1/2[mv^{2} ]$

therefore equating both the equations we get,

$qV=1/2[mv^{2} ]$ ⇒ $V=\frac{mv^{2} }{2q}$

m= mass of electron =  $9.1*10^{-31} kg$

q= charge on an electron = $1.6*10^{-19} C$

v= speed of electron= 700000m/s

substituting the values in the above equation, we get

$V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V$

(1).  the potential difference that stopped the electron is 1.39 volts.

now the kinetic energy equation is :  2 ways$KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J$

or $KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV$

(2).  the initial kinetic energy of the electron is 1.39eV.