The ground state energy of an atom is `-13.6eV`. The photon emitted during the transition of electron from `n=3` to `n=1` state, is incidenet on a photosensitive material of unknown work function. The photoelectrons are emitted from the materials with a maximum kinetic energy of 9 eV. the threshold wavelength of the material used is

Options

A. `0.9xx10^(-7)m`
B. `4xx10^(-7)m`
C. `0.47xx10^(-7)m`
D. `9xx10^(-7)m`

Explanation

To find the threshold wavelength of the photosensitive material, we will follow these steps:

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

The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]

For \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

Now, calculate the change in energy (\( \Delta E \)):
\[ \Delta E = E_3 - E_1 = (-1.51 \, \text{eV}) - (-13.6 \, \text{eV}) \]
\[ \Delta E = 12.09 \, \text{eV} \]

### Step 2: Use the photoelectric effect equation to find the work function.

According to the photoelectric effect, the energy of the photon (\( E \)) is related to the maximum kinetic energy of the emitted photoelectrons (\( K_{max} \)) and the work function (\( \phi \)) of the material:
\[ E = K_{max} + \phi \]

We know:
- \( K_{max} = 9 \, \text{eV} \)
- \( E = 12.09 \, \text{eV} \)

Now, rearranging the equation to find the work function:
\[ \phi = E - K_{max} \]
\[ \phi = 12.09 \, \text{eV} - 9 \, \text{eV} \]
\[ \phi = 3.09 \, \text{eV} \]

### Step 3: Calculate the threshold wavelength.

The threshold wavelength (\( \lambda_0 \)) can be calculated using the formula:
\[ \lambda_0 = \frac{hc}{\phi} \]

Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)
- \( \phi \) must be converted to joules:
\[ \phi = 3.09 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.944 \times 10^{-19} \, \text{J} \]

Now substituting the values:
\[ \lambda_0 = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{4.944 \times 10^{-19} \, \text{J}} \]
\[ \lambda_0 = \frac{1.9878 \times 10^{-25}}{4.944 \times 10^{-19}} \]
\[ \lambda_0 \approx 4.02 \times 10^{-7} \, \text{m} \]

### Final Result
The threshold wavelength of the material used is approximately \( 4.02 \times 10^{-7} \, \text{m} \) or \( 402 \, \text{nm} \).

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