
The work function of tungsten is $4.50$ $eV$ . Calculate the speed of the fastest electron ejected from the tungsten surface when light whose photon energy is $5.80$ $eV$ shines on the surface.
Answer
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Hint: The photoelectric effect defines the energy required to bring an electron from the inside of a metal to its surface, with the result that if the light incident has more energy than the work function, the electron comes out of the metal with additional kinetic energy. The difference between photon energy and work function is the maximum kinetic energy.
Formula used:
The general formula of kinetic energy which is,
$KE = \dfrac{1}{2}m{v^2}$
Complete step by step answer:
We are given that the value of the work function denoted by $W$ is $4.50$ $eV$ and the value of photon energy denoted by $E$ is $5.80$ $eV$. We know that the maximum kinetic energy will be the difference between the photon energy and work function.
Therefore, $KE = E - W$.
Now on putting the value i.e, $KE = 5.80 - 4.50$ that is equal to $1.3$ $eV$.
Now we will convert $eV$ into $Joule$, that is multiplying it by $1.6 \times {10^{ - 19}}$ $Joule$.
So it will become $1.3 \times 1.6 \times {10^{ - 19}}$ $Joule$.
Now apply the formula of kinetic energy that is $KE = \dfrac{1}{2}m{v^2}$.
We know that the value of mass $m$ is $9.1 \times {10^{ - 31}}$ $Kg$.
Now substitutes this value of mass and kinetic energy in the above formula.
Therefore, $1.3 \times 1.6 \times {10^{ - 19}} = \dfrac{{9.1 \times {{10}^{ - 31}} \times {v^2}}}{2}$
Now, ${v^2} = \dfrac{{1.3 \times 1.6 \times {{10}^{ - 19}} \times 2}}{{9.1 \times {{10}^{ - 31}}}}$
$\therefore v = 6.76 \times {10^5}$ $m/s$
Therefore our answer for the speed of the fastest electron ejected from the tungsten surface is $6.76 \times {10^5}$ $m/s$.
Note:One should remember the value of mass. Since all the energies are in electron volts so never forget to convert the value of $eV$ to $Joule$ . Always remember the formula while solving the problem. Check the units twice or thrice.Tungsten can survive to high temperature but its emission is limited due to its high work function.
Formula used:
The general formula of kinetic energy which is,
$KE = \dfrac{1}{2}m{v^2}$
Complete step by step answer:
We are given that the value of the work function denoted by $W$ is $4.50$ $eV$ and the value of photon energy denoted by $E$ is $5.80$ $eV$. We know that the maximum kinetic energy will be the difference between the photon energy and work function.
Therefore, $KE = E - W$.
Now on putting the value i.e, $KE = 5.80 - 4.50$ that is equal to $1.3$ $eV$.
Now we will convert $eV$ into $Joule$, that is multiplying it by $1.6 \times {10^{ - 19}}$ $Joule$.
So it will become $1.3 \times 1.6 \times {10^{ - 19}}$ $Joule$.
Now apply the formula of kinetic energy that is $KE = \dfrac{1}{2}m{v^2}$.
We know that the value of mass $m$ is $9.1 \times {10^{ - 31}}$ $Kg$.
Now substitutes this value of mass and kinetic energy in the above formula.
Therefore, $1.3 \times 1.6 \times {10^{ - 19}} = \dfrac{{9.1 \times {{10}^{ - 31}} \times {v^2}}}{2}$
Now, ${v^2} = \dfrac{{1.3 \times 1.6 \times {{10}^{ - 19}} \times 2}}{{9.1 \times {{10}^{ - 31}}}}$
$\therefore v = 6.76 \times {10^5}$ $m/s$
Therefore our answer for the speed of the fastest electron ejected from the tungsten surface is $6.76 \times {10^5}$ $m/s$.
Note:One should remember the value of mass. Since all the energies are in electron volts so never forget to convert the value of $eV$ to $Joule$ . Always remember the formula while solving the problem. Check the units twice or thrice.Tungsten can survive to high temperature but its emission is limited due to its high work function.
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