Answer

Verified

414.9k+ views

**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:**

**Complete step by step answer:**

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.

Recently Updated Pages

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE