 The threshold frequency for a certain metal is $3.3\times {{10}^{14}}cycle/sec$. If the incident light on the metal has a cut-off frequency of $8.2\times {{10}^{14}}cycle/sec$. Calculate the cut-off potential for the photoelectron.A. $2 volt$B. $0.5 volt$C. $1 volt$D. $4 volt$ Verified
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Hint: Recall Einstein’s photoelectric equation and how it relates the kinetic energy of the photoelectron, the work function and the energy of a photo. Relate these terms with the Planck’s constant and simplify to find the relation between the threshold frequency, cut-off frequency, and the potential.

Formula used:

We know that Einstein’s photoelectric formula says that the energy of a photon of light is equal to the sum of the work function and the maximum kinetic energy of the photoelectron. The formula is:
$E=\phi +{{E}_{k}}$
Where, $E$ is the energy of the photon, $\phi$ is the work function, and ${{E}_{k}}$ is the kinetic energy of the photoelectron. Here, we can see that, if the photons from the incident light do not have a frequency that is enough to overcome the energy barrier of the work function, then no electrons will be emitted. Thus, $E$ corresponds to the energy of the incident light, $\phi$ corresponds to the threshold energy required to displace electrons, and ${{E}_{k}}$ refers to the kinetic energy of the photoelectrons after they are emitted.
We know that, according to the De Broglie equation; $E = hv$.
Where, $E$ is the energy of the photon, $h$ s the Planck’s constant, and $v$ is the frequency of the photons.
The threshold energy which is defined as the work function can also be translated as $\phi = h{{v}_{o}}$.
Where, ${{v}_{o}}$ is the threshold frequency.
We know that the kinetic energy of an electron can be defined as the product of the charge on an electron and its cut-off potential. Thus, we can rearrange Einstein’s photoelectric equation as:
$hv=h{{v}_{o}}+eV$
Where, $e$ is the charge on an electron and $V$ is the cut-off potential. Rearranging the equation to find the cut-off potential, we get:
$V=\frac{1}{e}(hv-h{{v}_{o}})$
We already know from the given information that:
Charge on electron = $e =1.6\times {{10}^{-19}}C$
Planck’s constant = $h = 6.6\times {{10}^{-34}}J$
Frequency of incident light = $v = 8.2\times {{10}^{14}}Hz$
Threshold frequency = ${{v}_{o}} = 3.3\times {{10}^{14}}$
Thus, putting the values in the equation, we get:
$V=\frac{6.6\times {{10}^{-34}}}{1.6\times {{10}^{-19}}}\times (8.2-3.3)\times {{10}^{14}}volt$
Here, we have taken Planck's constant and the order of magnitude of the frequency (${{10}^{14}}$) as common. Solving the equation further, we get:
\begin{align} & V = 4.125\times 4.9\times {{10}^{-1}}volt \\ & V = 20.2125\times {{10}^{-1}}volt \\ & V = 2.02 volt \\ \end{align}
Thus, the cut-off potential comes out to be 2.02volts. Rounding it off to get one of the options given,

the answer to this question is ‘A. $2 volt$’.

Note: Remember that the $eV$ given here does not mean electron volts. It indicates the charge present on an electron and the cut-off potential of a photoelectron. Please do not confuse the two concepts as the electron volt is a unit of potential.