Surface of sodium is illuminated by a light of 6000 Å wavelength. Work function of sodium is 2.6 eV. Then minimum K.E. of emitted electron is :

Options

A. 0 eV
B. 1.53 eV
C. 2.46 eV
D. 4.14 eV

Answer

To solve the problem step by step, we need to find the minimum kinetic energy of the emitted electrons when sodium is illuminated by light of a specific wavelength. Here’s how we can approach this:

### Step 1: Understand the Given Information
- Wavelength of light (\( \lambda \)) = 6000 Å = 6000 x \( 10^{-10} \) m = 600 nm
- Work function of sodium (\( \phi \)) = 2.6 eV

### Step 2: Calculate the Energy of the Incident Photons
The energy of the incident photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \) J·s
- \( c \) (speed of light) = \( 3 \times 10^8 \) m/s
- \( \lambda \) = 600 nm = \( 600 \times 10^{-9} \) m

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \text{ J·s})(3 \times 10^8 \text{ m/s})}{600 \times 10^{-9} \text{ m}} \]

Calculating this gives:
\[ E = \frac{1.9878 \times 10^{-25}}{600 \times 10^{-9}} = 3.313 \times 10^{-19} \text{ J} \]

### Step 3: Convert Energy from Joules to Electron Volts
To convert energy from Joules to electron volts, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ E = \frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.07 \text{ eV} \]

### Step 4: Compare the Energy of Photons with the Work Function
Now, we compare the energy of the photons (\( E \)) with the work function (\( \phi \)):
- \( E \approx 2.07 \text{ eV} \)
- \( \phi = 2.6 \text{ eV} \)

Since \( E < \phi \), the energy of the incident photons is less than the work function of sodium.

### Step 5: Determine the Kinetic Energy of Emitted Electrons
According to the photoelectric effect, if the energy of the incident photons is less than the work function, no electrons will be emitted. Therefore, the minimum kinetic energy of the emitted electrons is:
\[ \text{K.E.} = 0 \text{ eV} \]

### Final Answer
The minimum kinetic energy of the emitted electrons is **0 eV**.

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