If the work function `(w)` of an arbitrary metal is `3.1 eV`, find its threshold wavelength and the maximum kinetic energy of the electron emitted when radiation of `300 nm` strike the metal surface. `("Take" hc=12400 eV Å)`

Answer

To solve the problem, we need to find two things: the threshold wavelength and the maximum kinetic energy of the emitted electron when radiation of 300 nm strikes the metal surface.

### Step 1: Calculate the Threshold Wavelength

The threshold wavelength (λ₀) can be calculated using the formula:

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

Where:
- \( h \) is Planck's constant multiplied by the speed of light (given as \( 12400 \, \text{eV} \cdot \text{Å} \))
- \( c \) is the speed of light (already included in the value of \( hc \))
- \( W \) is the work function of the metal (given as \( 3.1 \, \text{eV} \))

Substituting the values:

\[ \lambda_0 = \frac{12400 \, \text{eV} \cdot \text{Å}}{3.1 \, \text{eV}} = 4000 \, \text{Å} \]

### Step 2: Calculate the Energy of the Incident Photon

Next, we need to calculate the energy of the photon (E) when radiation of 300 nm strikes the metal surface. We can use the formula:

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

Where:
- \( \lambda \) is the wavelength of the incident radiation (300 nm = \( 3000 \, \text{Å} \))

Substituting the values:

\[ E = \frac{12400 \, \text{eV} \cdot \text{Å}}{300 \, \text{nm}} = \frac{12400 \, \text{eV} \cdot \text{Å}}{3000 \, \text{Å}} = 4.133 \, \text{eV} \]

### Step 3: Calculate the Maximum Kinetic Energy of the Emitted Electron

The maximum kinetic energy (KE_max) of the emitted electron can be calculated using the formula:

\[ KE_{max} = E - W \]

Where:
- \( E \) is the energy of the incident photon (calculated in Step 2)
- \( W \) is the work function (given as \( 3.1 \, \text{eV} \))

Substituting the values:

\[ KE_{max} = 4.133 \, \text{eV} - 3.1 \, \text{eV} = 1.033 \, \text{eV} \]

### Final Answers

- Threshold Wavelength: \( 4000 \, \text{Å} \)
- Maximum Kinetic Energy: \( 1.033 \, \text{eV} \)

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