The electron in the hydrogen atom jumps from excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1eV, the stopping potential is estimated to be: (The energy of the electron in nth state is `E_(n)=-13.6//n^(2)eV`)

Options

A. `12.1 V`
B. `17.2 V`
C. 7 V
D. `5.1 V`

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the energy of the photon emitted during the transition from n=3 to n=1.

The energy of the electron in the nth state of a hydrogen atom is given by the formula:

\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]

For \( n = 3 \):

\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \text{ eV} \approx -1.51 \text{ eV} \]

For \( n = 1 \):

\[ E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \]

Now, the energy of the photon emitted when the electron transitions from n=3 to n=1 is:

\[ E = E_1 - E_3 = -13.6 \text{ eV} - (-1.51 \text{ eV}) = -13.6 + 1.51 = -12.1 \text{ eV} \]

Thus, the energy of the emitted photon is:

\[ E = 12.1 \text{ eV} \]

### Step 2: Relate the energy of the emitted photon to the work function and kinetic energy.

The energy of the emitted photon can be expressed in terms of the work function (W) and the kinetic energy (KE) of the emitted electrons:

\[ E = KE + W \]

Given that the work function \( W = 5.1 \text{ eV} \), we can rearrange the equation to solve for kinetic energy:

\[ KE = E - W = 12.1 \text{ eV} - 5.1 \text{ eV} = 7.0 \text{ eV} \]

### Step 3: Calculate the stopping potential.

The kinetic energy of the emitted electrons can also be expressed in terms of the stopping potential (V):

\[ KE = eV \]

Where \( e \) is the charge of the electron (1 eV corresponds to the energy gained by an electron when it is accelerated through a potential difference of 1 volt). Thus, we can set:

\[ 7.0 \text{ eV} = eV \]

From this, we can find the stopping potential:

\[ V = 7.0 \text{ V} \]

### Final Answer:

The stopping potential is:

\[ \text{Stopping Potential} = 7.0 \text{ V} \]

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