In photoelectric effect, the threshold wavelength of sodium is $5000\overset{o}{\mathop{A}}\,$. Find its work function. $(h=6.6\times {{10}^{-34}}Js,c=3\times {{10}^{8}}m{{s}^{-1}},1eV=1.6\times {{10}^{-19}}J)$
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Hint: In photoelectric effect, work function of a metal surface is easily calculated using the energy equation of photons. The answer determined in this manner is usually in joule. To match the answer given in options, it is important to convert the answer from $joule$ to $eV$ using the given conversion formula.
Formula used:
$1)W=\dfrac{hc}{\lambda }$
where
$W$ is the work function of the substance
$h$ is the Planck’s constant
$c$ is the speed of light
$\lambda $ is the threshold frequency of the substance
$2)1eV=1.6\times {{10}^{-19}}J$
Complete step by step answer:
Photoelectric effect is the phenomenon of emission of electrons when light (photons) strikes the surface of a metal. Work function in photoelectric effect is the minimum energy of photons required to liberate an electron from the surface of the metal. For photoelectric effect to happen, the energy of the photon should be greater than the work function of the metal.
Work function of a metal is given by
$W=\dfrac{hc}{\lambda }$
Given
$\begin{align}
& h=6.6\times {{10}^{-34}}Js \\
& c=3\times {{10}^{8}}m{{s}^{-1}} \\
& \lambda =5000\overset{o}{\mathop{A}}\,=5000\times {{10}^{-10}}m \\
\end{align}$
Substituting these values in the above equation,
$W=\dfrac{hc}{\lambda }=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}=3.96\times {{10}^{-19}}J$
Now, let us convert this energy into $eV$. We know that
$\begin{align}
& 1eV=1.6\times {{10}^{-19}}J \\
& 1J=\dfrac{1}{1.6\times {{10}^{-19}}}eV \\
& \\
\end{align}$
So,
$3.96\times {{10}^{-19}}J=\dfrac{3.96\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV=2.475eV\approx 2.5eV$
So, the correct answer is “Option B”.
Additional Information: For a metal surface, threshold frequency is the minimum frequency of incident radiation required for photoelectric effect to happen. If the incident radiation has a frequency below the threshold frequency of the metal surface, no photoelectrons will be emitted. It is usually represented by $\lambda $ or ${{\lambda }_{0}}$.
Note: Work function of a metal is sometimes represented by $\Psi $. Students need not get confused with the symbol or the letter used for parameters in such questions. What needs to be concentrated more is the correct relation or formula. Students need to know the conversion formulas. The value of Planck’s constant should also be learnt thoroughly. It is not necessary that all the values and conversion formulas will be provided in the question always.
Formula used:
$1)W=\dfrac{hc}{\lambda }$
where
$W$ is the work function of the substance
$h$ is the Planck’s constant
$c$ is the speed of light
$\lambda $ is the threshold frequency of the substance
$2)1eV=1.6\times {{10}^{-19}}J$
Complete step by step answer:
Photoelectric effect is the phenomenon of emission of electrons when light (photons) strikes the surface of a metal. Work function in photoelectric effect is the minimum energy of photons required to liberate an electron from the surface of the metal. For photoelectric effect to happen, the energy of the photon should be greater than the work function of the metal.
Work function of a metal is given by
$W=\dfrac{hc}{\lambda }$
Given
$\begin{align}
& h=6.6\times {{10}^{-34}}Js \\
& c=3\times {{10}^{8}}m{{s}^{-1}} \\
& \lambda =5000\overset{o}{\mathop{A}}\,=5000\times {{10}^{-10}}m \\
\end{align}$
Substituting these values in the above equation,
$W=\dfrac{hc}{\lambda }=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}=3.96\times {{10}^{-19}}J$
Now, let us convert this energy into $eV$. We know that
$\begin{align}
& 1eV=1.6\times {{10}^{-19}}J \\
& 1J=\dfrac{1}{1.6\times {{10}^{-19}}}eV \\
& \\
\end{align}$
So,
$3.96\times {{10}^{-19}}J=\dfrac{3.96\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV=2.475eV\approx 2.5eV$
So, the correct answer is “Option B”.
Additional Information: For a metal surface, threshold frequency is the minimum frequency of incident radiation required for photoelectric effect to happen. If the incident radiation has a frequency below the threshold frequency of the metal surface, no photoelectrons will be emitted. It is usually represented by $\lambda $ or ${{\lambda }_{0}}$.
Note: Work function of a metal is sometimes represented by $\Psi $. Students need not get confused with the symbol or the letter used for parameters in such questions. What needs to be concentrated more is the correct relation or formula. Students need to know the conversion formulas. The value of Planck’s constant should also be learnt thoroughly. It is not necessary that all the values and conversion formulas will be provided in the question always.