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 :
Given E (in eV)`=(1237)/(lambda("in nm"))`

Options

A. 1.5 eV
B. 3.0 eV
C. 4.5 eV
D. 15.1 eV

Answer

To solve the problem, we will use the formula for the maximum kinetic energy of emitted electrons in the photoelectric effect. The formula is given by:

\[ K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]

Where:
- \( K.E. \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \lambda_0 \) is the threshold wavelength.

### Step 1: Identify the given values
- Threshold wavelength (\( \lambda_0 \)) = 380 nm
- Incident wavelength (\( \lambda \)) = 260 nm

### Step 2: Convert the formula
We can factor out \( hc \) from the equation:

\[ K.E. = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

### Step 3: Substitute the values of \( \lambda \) and \( \lambda_0 \)
Now we substitute the values into the equation:

\[ K.E. = hc \left( \frac{1}{260} - \frac{1}{380} \right) \]

### Step 4: Calculate \( \frac{1}{\lambda} - \frac{1}{\lambda_0} \)
Calculate \( \frac{1}{260} - \frac{1}{380} \):

1. Find a common denominator, which is \( 260 \times 380 \).
2. Rewrite the fractions:
\[ \frac{1}{260} = \frac{380}{260 \times 380} \]
\[ \frac{1}{380} = \frac{260}{260 \times 380} \]
3. Now, subtract:
\[ \frac{380 - 260}{260 \times 380} = \frac{120}{260 \times 380} \]

### Step 5: Substitute back into the kinetic energy equation
Now substitute this back into the kinetic energy formula:

\[ K.E. = hc \left( \frac{120}{260 \times 380} \right) \]

### Step 6: Use the value of \( hc \)
Given \( hc = 1237 \) eV·nm, we can substitute this value:

\[ K.E. = 1237 \left( \frac{120}{260 \times 380} \right) \]

### Step 7: Calculate the kinetic energy
Now we calculate:

1. Calculate \( 260 \times 380 = 98800 \).
2. Now substitute:
\[ K.E. = 1237 \left( \frac{120}{98800} \right) \]
3. Calculate \( \frac{120}{98800} = 0.001214 \).
4. Finally, calculate:
\[ K.E. = 1237 \times 0.001214 \approx 1.5 \text{ eV} \]

### Conclusion
Thus, the maximum kinetic energy of the emitted electrons is approximately \( 1.5 \text{ eV} \).

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