In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be ________eV
Given E (in eV) = `(1237)/(lambda ("in nm"))`

Explanation

To solve the problem of finding the maximum kinetic energy of emitted electrons in a photoelectric effect experiment, we can follow these steps:

### Step 1: Understand the given data
- Threshold wavelength (λ₀) = 380 nm
- Wavelength of incident light (λ) = 260 nm

### Step 2: Calculate the energy of the incident photon
We can use the formula for the energy of a photon in electron volts (eV):
\[ E = \frac{1237}{\lambda \text{ (in nm)}} \]
Substituting the wavelength of the incident light:
\[ E_{\text{incident}} = \frac{1237}{260} \approx 4.75 \text{ eV} \]

### Step 3: Calculate the work function (φ)
The work function can be calculated using the threshold wavelength:
\[ E_{\text{threshold}} = \frac{1237}{\lambda_0 \text{ (in nm)}} \]
Substituting the threshold wavelength:
\[ E_{\text{threshold}} = \frac{1237}{380} \approx 3.25 \text{ eV} \]

### Step 4: Calculate the maximum kinetic energy (K.E.)
Using Einstein's photoelectric equation:
\[ K.E. = E_{\text{incident}} - E_{\text{threshold}} \]
Substituting the values we calculated:
\[ K.E. = 4.75 \text{ eV} - 3.25 \text{ eV} = 1.5 \text{ eV} \]

### Final Answer
The maximum kinetic energy of emitted electrons will be **1.5 eV**.

---