Ligth of wavelength `4000A^(@)` is incident on a metal surface of work function `2.5 eV`. Given `h=6.62xx10^(-34) Js,c=3xx10^(8) m//s`, the maximum `KE` of photoelectrons emitted and the corresponding stopping potential are respectively

Options

A. `0.6 eV,0.6 V`
B. `2.5 eV,2.5 V`
C. `3.1 eV,3.1 V`
D. `0.6 eV,0.3 V`

Explanation

To solve the problem step by step, we will calculate the maximum kinetic energy (KE) of the photoelectrons emitted and the corresponding stopping potential.

### Step 1: Calculate the Energy of the Incident Light
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.62 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^{8} \, \text{m/s} \) (speed of light)
- \( \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \) (wavelength of light)

Substituting the values:
\[ E = \frac{(6.62 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{4000 \times 10^{-10} \, \text{m}} \]

Calculating the energy:
\[ E = \frac{1.986 \times 10^{-25} \, \text{Js}}{4000 \times 10^{-10} \, \text{m}} = 4.965 \times 10^{-19} \, \text{J} \]

To convert this energy from Joules to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E = \frac{4.965 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 3.10 \, \text{eV} \]

### Step 2: Calculate the Maximum Kinetic Energy of Photoelectrons
The maximum kinetic energy (KE) of the emitted photoelectrons is given by:
\[ KE_{\text{max}} = E - \phi \]
where \( \phi \) is the work function of the metal, given as \( 2.5 \, \text{eV} \).

Substituting the values:
\[ KE_{\text{max}} = 3.10 \, \text{eV} - 2.5 \, \text{eV} = 0.60 \, \text{eV} \]

### Step 3: Calculate the Stopping Potential
The stopping potential \( V_0 \) can be calculated using the relationship:
\[ KE_{\text{max}} = eV_0 \]
where \( e \) is the charge of an electron (1 eV corresponds to 1 volt for an electron).

Thus,
\[ V_0 = KE_{\text{max}} = 0.60 \, \text{V} \]

### Final Results
- The maximum kinetic energy of the photoelectrons emitted is \( 0.60 \, \text{eV} \).
- The corresponding stopping potential is \( 0.60 \, \text{V} \).