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 solve the problem step by step, we need to find the threshold wavelength of the photosensitive material based on the given information. Here's how we can approach it:

### Step 1: Calculate the Energy of the Photon Emitted
The energy of the photon emitted during the transition from \( n=3 \) to \( n=1 \) can be calculated using the formula for the energy levels of hydrogen-like atoms:

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

For the transition from \( n=3 \) to \( n=1 \):

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

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

The energy of the emitted photon is given by:

\[ E_{\text{photon}} = E_{1} - E_{3} = (-13.6) - (-1.51) = -13.6 + 1.51 = -12.09 \, \text{eV} \]

### Step 2: Calculate the Maximum Kinetic Energy of the Photoelectrons
The maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectrons is given as 9 eV. According to the photoelectric effect equation:

\[ K_{\text{max}} = E_{\text{photon}} - \phi \]

Where \( \phi \) is the work function of the material. Rearranging gives us:

\[ \phi = E_{\text{photon}} - K_{\text{max}} = 12.09 \, \text{eV} - 9 \, \text{eV} = 3.09 \, \text{eV} \]

### Step 3: Calculate the Threshold Wavelength
The threshold wavelength \( \lambda_0 \) can be calculated using the relation between energy and wavelength:

\[ E = \frac{hc}{\lambda} \]

Rearranging gives:

\[ \lambda_0 = \frac{hc}{\phi} \]

Substituting the values of \( h \) (Planck's constant) and \( c \) (speed of light) into the equation:

\[ h = 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} \]
\[ c = 3 \times 10^{8} \, \text{m/s} \]

Thus,

\[ \lambda_0 = \frac{(4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s})(3 \times 10^{8} \, \text{m/s})}{3.09 \, \text{eV}} \]

Calculating this gives:

\[ \lambda_0 \approx \frac{1.241 \times 10^{-6} \, \text{eV}\cdot\text{m}}{3.09 \, \text{eV}} \approx 4.02 \times 10^{-7} \, \text{m} \approx 4.02 \times 10^{-7} \, \text{m} \]

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

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