Light of wavelength `5000 Å` falls on a sensitive plate with photoelectric work function of `1.9 eV`. The kinetic energy of the photoelectron emitted will be

Options

A. `0.58 eV`
B. `2.48 eV`
C. `1.24 eV`
D. `1.16 eV`

Explanation

To find the kinetic energy of the photoelectron emitted when light of wavelength `5000 Å` falls on a sensitive plate with a work function of `1.9 eV`, we can follow these steps:

### Step 1: Convert the wavelength from angstroms to meters
1 angstrom (Å) = \(10^{-10}\) meters. Therefore,
\[ \text{Wavelength} = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate the energy of the incident photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \(h\) (Planck's constant) = \(4.1357 \times 10^{-15} \, \text{eV s}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda\) = \(5 \times 10^{-7} \, \text{m}\)

Now substituting the values:
\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV s}) \times (3 \times 10^8 \, \text{m/s})}{5 \times 10^{-7} \, \text{m}} \]

Calculating this gives:
\[ E = \frac{1.24071 \times 10^{-6} \, \text{eV m}}{5 \times 10^{-7} \, \text{m}} = 2.48142 \, \text{eV} \]

### Step 3: Calculate the kinetic energy of the emitted photoelectron
Using Einstein's photoelectric equation:
\[ K_{max} = E - \phi \]
Where:
- \(E\) = energy of the incident photon = \(2.48142 \, \text{eV}\)
- \(\phi\) = work function = \(1.9 \, \text{eV}\)

Substituting the values:
\[ K_{max} = 2.48142 \, \text{eV} - 1.9 \, \text{eV} = 0.58142 \, \text{eV} \]

### Final Answer
The kinetic energy of the photoelectron emitted will be approximately \(0.58 \, \text{eV}\).

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