A light of wavelength `5000 Å` falls on a photosensitive plate with photoelectric work function 1.90 eV. Kinetic energy of the emitted photoelectrons will be `(h = 6.63 xx 10^(-34)J.s)`

Options

A. 0.1 eV
B. 0.59 eV
C. 1.581 eV
D. 2 eV

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the wavelength from Angstroms to meters (if needed)
The given wavelength is \( \lambda = 5000 \, \text{Å} \).
1 Ångström = \( 10^{-10} \, \text{m} \).
So,
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m}. \]

### Step 2: Calculate the energy of the incident photons
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h = 6.63 \times 10^{-34} \, \text{J.s} \) and \( c = 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J.s}) \times (3 \times 10^8 \, \text{m/s})}{5 \times 10^{-7} \, \text{m}}. \]
Calculating this gives:
\[ E = \frac{1.989 \times 10^{-25} \, \text{J.m}}{5 \times 10^{-7} \, \text{m}} = 3.978 \times 10^{-19} \, \text{J}. \]

### Step 3: Convert the energy from Joules to electron volts
1 eV = \( 1.6 \times 10^{-19} \, \text{J} \).
Thus,
\[ E = \frac{3.978 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.49 \, \text{eV}. \]

### Step 4: Calculate the maximum kinetic energy of the emitted photoelectrons
The maximum kinetic energy (KE) of the emitted photoelectrons can be calculated using the equation:
\[ KE_{\text{max}} = E - \phi \]
where \( \phi = 1.90 \, \text{eV} \).

Substituting the values:
\[ KE_{\text{max}} = 2.49 \, \text{eV} - 1.90 \, \text{eV} = 0.59 \, \text{eV}. \]

### Final Answer
The kinetic energy of the emitted photoelectrons is \( \boxed{0.59 \, \text{eV}} \).

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