Light of frequency `8 xx 10^(15) Hz` is incident on a substance of photoelectric work function `6.125 eV`. The maximum kinetic energy of the emitted photoelectrons is

Options

A. `17 eV`
B. `22 eV`
C. `27 eV`
D. `37 eV`

Explanation

To find the maximum kinetic energy of the emitted photoelectrons when light of frequency \(8 \times 10^{15} \, \text{Hz}\) is incident on a substance with a photoelectric work function of \(6.125 \, \text{eV}\), we can follow these steps:

### Step 1: Calculate the energy of the incident photons
The energy \(E\) of a photon can be calculated using the formula:
\[ E = h \nu \]
where:
- \(h\) is Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- \(\nu\) is the frequency of the light, \(\nu = 8 \times 10^{15} \, \text{Hz}\)

Substituting the values:
\[ E = (6.626 \times 10^{-34} \, \text{J s}) \times (8 \times 10^{15} \, \text{Hz}) \]

### Step 2: Calculate the energy in joules
Calculating the above expression:
\[ E = 6.626 \times 8 \times 10^{-34 + 15} = 53.008 \times 10^{-19} \, \text{J} \]

### Step 3: Convert the energy from joules to electron volts
To convert joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E \, (\text{in eV}) = \frac{53.008 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 33.13 \, \text{eV} \]

### Step 4: Calculate the maximum kinetic energy of the emitted photoelectrons
The maximum kinetic energy \(K_{\text{max}}\) of the emitted photoelectrons is given by:
\[ K_{\text{max}} = E - \phi \]
where \(\phi\) is the work function of the substance.

Substituting the values:
\[ K_{\text{max}} = 33.13 \, \text{eV} - 6.125 \, \text{eV} \approx 26.995 \, \text{eV} \]

### Step 5: Round off the answer
Rounding off gives us:
\[ K_{\text{max}} \approx 27 \, \text{eV} \]

Thus, the maximum kinetic energy of the emitted photoelectrons is approximately \(27 \, \text{eV}\).