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Question: The K.E. of an electron emitted from a tungsten surface is \[3\cdot 06\text{ }eV\]. What voltage wou...

The K.E. of an electron emitted from a tungsten surface is 3⋅06 eV3\cdot 06\text{ }eV. What voltage would be required to bring the electron to rest?

Explanation

Solution

Convert the kinetic energy into joules (1 eV= 1⋅6×10−19J)\left( 1\text{ }eV=~1\cdot 6\times {{10}^{-19}}J \right) and then use the formula of work done which equals the kinetic energy. Equating both we will get the required value of voltage to bring the electron to rest.
Formula Used:
qV=12mv2qV=\dfrac{1}{2}m{{v}^{2}}
Where, q=q= Charge on electron
V=V= Potential difference
m=m= Mass of electron
v=v= Velocity of electron
12mv2=\dfrac{1}{2}m{{v}^{2}}= Kinetic energy of electron

Complete Step by step Solution:
Given K.E. of electron =3⋅06 eV=3\cdot 06\text{ }eV
Converting eV into Joules we get, K.E.= 3⋅06×1⋅6×10−19J\text{= }3\cdot 06\times 1\cdot 6\times {{10}^{-19}}\text{J}
Now using the above formula, qV=12mv2qV=\dfrac{1}{2}m{{v}^{2}}
qV=3⋅06×1⋅6×10−19qV=3\cdot 06\times 1\cdot 6\times {{10}^{-19}} J
Charge on electron, (q)=1⋅6×10−19C\left( q \right)=1\cdot 6\times {{10}^{-19}}C
So, (1⋅6×10−19C)V=3⋅06×1⋅6×10−19C\left( 1\cdot 6\times {{10}^{-19}}C \right)V=3\cdot 06\times 1\cdot 6\times {{10}^{-19}}C
V=3⋅06×1⋅6×10−19J1⋅6×10−19CV=\dfrac{3\cdot 06\times 1\cdot 6\times {{10}^{-19}}J}{1\cdot 6\times {{10}^{-19}}C}
=3⋅06JC−1=3\cdot 06J{{C}^{-1}}
=3⋅06 V=3\cdot 06\text{ }V

So, 3.06 V3.06\text{ }V is required to bring the electron to rest.

Additional Information:
The S.I unit of energy is Joule. Another convenient unit of energy is electron volt (eV)\left( eV \right) which is the kinetic energy gained or lost by an electron in moving through a potential difference of one volt.
1electron volt
=1 eV=1⋅6×10−19C×1 V=1⋅6×10−19J=1\text{ }eV=1\cdot 6\times {{10}^{-19}}C\times 1\text{ }V=1\cdot 6\times {{10}^{-19}}J
Electrostatic Potential: At any point in the region of electrostatic field it is the minimum work done in carrying a unit positive charge (without acceleration) from infinity to that point.
WABq=VB−VA=ΔV\dfrac{{{W}_{AB}}}{q}={{V}_{B}}-{{V}_{A}}=\Delta V
SI unit of electrostatic potential difference is volt. That is why potential difference is referred to as voltage.
1V=1J1C=1JC−1=1NmC−11V=\dfrac{1J}{1C}=1J{{C}^{-1}}=1Nm{{C}^{-1}}
Kinetic Energy: It is the energy possessed by an object due to its motion. It is defined as the work needed to accelerate a body of given mass from rest to its stated velocity. It is produced by the movement of the body. It increases with speed. It is higher if the mass of the body that produces it is greater.
It is an important tool in the macroscopic world. It is one of the ways in which we relate mass and motion and force to describe the state of an object and predicts its interactions with the rest of the world. The kinetic energy of an object depends on both its mass and speed. It increases as mass and speed are increased.

Note: Study the formula of kinetic energy, potential difference and their relation to one another. One can study the Photoelectric effect which will explain the emission of electrons from the surface of tungsten metal with kinetic energy. It is the effect in which electrons are emitted from a material that has absorbed electromagnetic radiation.