Respuesta :
As we know that electrostatic potential difference is used to stop the moving electron
So here kinetic energy of electron is converted into electrostatic potential energy
so we will first find the kinetic energy of moving electron as
[tex]KE = \frac{1}{2}mv^2[/tex]
now we have
[tex]m = 9.1 \times 10^{-31} kg[/tex]
[tex]v = 6.00 \times 10^5 m/s[/tex]
now from above formula we have
[tex]KE = \frac{1}{2}(9.1 \times 10^{-31})(6.00 \times 10^5)^2[/tex]
[tex]KE = 1.64 \times 10^{-19} J[/tex]
now to convert kinetic energy into electron volt we know that
[tex]1eV = 1.6 \times 10^{-19} J[/tex]
so we have
[tex]KE = 1.02 eV[/tex]
now from energy conservation change in electrostatic potential energy is given as
[tex]\Delta U = \Delta KE[/tex]
[tex]q(\Delta V) = 1.64 \times 10^{-19}[/tex]
[tex]1.6 \times 10^{-19}(\Delta V) = 1.64 \times 10^{-19}[/tex]
[tex]\Delta V = 1.02 Volts[/tex]
so it is stopped by 1.02 Volts potential difference
The electron which brought to rest by an electric field has,
- The potential difference that stopped the electron is 1.02 volts.
- The initial kinetic energy of the electron, in electron volts is 1.02 eV.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The initial speed of the electron is [tex]6.00\times 10^{5}[/tex] meter per second.
The electron is brought to the rest by electric field.
- a) The potential difference that stopped the electron-
As the mass (m) of a electron is [tex]9.1\times10{-31}[/tex] kg. Thus kinetic energy (KE) of it can be given as,
[tex]KE=\dfrac{1}{2}mv^2[/tex]
[tex]KE=\dfrac{1}{2}\times9.1\times10^{-31}\times(6.00\times10^{5})^2\\KE=1.64\times10^{-19} \rm J[/tex]
Now this kinetic energy should be equal to the potential energy, which stopped the electron. The charge (q) of a electron is [tex]1.6\times10^{-19}[/tex] C. Thus potential energy
[tex]q\Delta V=1.64\times10^{-19}\\1.6\times10^{-19}\Delta V=1.64\times10^{-19}\\\Delta V=1.02\rm Volts[/tex]
Thus, the potential difference that stopped the electron is 1.02 volts.
- b) The initial kinetic energy of the electron, in electron volts-
As the mass (m) of a electron is [tex]9.1\times10{-31}[/tex] kg. Thus kinetic energy (KE) of it can be given as,
[tex]KE=\dfrac{1}{2}mv^2[/tex]
[tex]KE=\dfrac{1}{2}\times9.1\times10^{-31}\times(6.00\times10^{5})^2\\KE=1.64\times10^{-19} \rm J[/tex]
As the charge (q) of a electron 1 ev is [tex]1.6\times10^{-19}[/tex] C. Thus the kinetic energy of electron is,
[tex]KE=1.02eV[/tex]
Thus, the initial kinetic energy of the electron, in electron volts is 1.02 eV.
Hence,
- The potential difference that stopped the electron is 1.02 volts.
- The initial kinetic energy of the electron, in electron volts is 1.02 eV.
Learn more about electric field here;
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