When a certain metal was irradiated with a light of `8.1 xx 10^(16)Hz` frequency, the photoelectron emitted had 1.5 times the kinetic energy as the photoelectrons emitted when the same metal was irradiated with light `5.8 xx 10^(16)Hz` frequency. If the same metal is irradiated with light of `3.846nm` wave length, what will be the energy of the photoelectron emitted?

Options

A. `1.8 xx 10^(2)eV`
B. `3.65 xx 10^(-17)J`
C. `2.28 xx 10^(2)eV`
D. `4.37 xx 10^(-17)J`

Answer

To solve the problem, we need to determine the energy of the photoelectron emitted when the metal is irradiated with light of wavelength 3.846 nm. We will use the photoelectric effect principles, which relate the energy of the incident light to the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Calculate the energy of the incident light for both frequencies.**
The energy (E) of a photon can be calculated using the formula:
\[ E = h \cdot f \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)) and \( f \) is the frequency.

- For \( f_1 = 8.1 \times 10^{16} \, \text{Hz} \):
\[ E_1 = h \cdot f_1 = 6.626 \times 10^{-34} \cdot 8.1 \times 10^{16} \]
\[ E_1 \approx 5.37 \times 10^{-17} \, \text{J} \]

- For \( f_2 = 5.8 \times 10^{16} \, \text{Hz} \):
\[ E_2 = h \cdot f_2 = 6.626 \times 10^{-34} \cdot 5.8 \times 10^{16} \]
\[ E_2 \approx 3.84 \times 10^{-17} \, \text{J} \]

2. **Relate the kinetic energies of the emitted photoelectrons.**
According to the problem, the kinetic energy of the photoelectrons emitted when irradiated with \( f_1 \) is 1.5 times that of those emitted with \( f_2 \):
\[ KE_1 = 1.5 \times KE_2 \]
The kinetic energy of the photoelectrons can be expressed as:
\[ KE = E - \phi \]
where \( \phi \) is the work function of the metal.

Therefore, we can write:
\[ E_1 - \phi = 1.5 \times (E_2 - \phi) \]

3. **Substitute the values of \( E_1 \) and \( E_2 \) into the equation.**
\[ 5.37 \times 10^{-17} - \phi = 1.5 \times (3.84 \times 10^{-17} - \phi) \]
Expanding the right side:
\[ 5.37 \times 10^{-17} - \phi = 5.76 \times 10^{-17} - 1.5\phi \]
Rearranging gives:
\[ 5.37 \times 10^{-17} + 1.5\phi = 5.76 \times 10^{-17} + \phi \]
\[ 0.5\phi = 5.76 \times 10^{-17} - 5.37 \times 10^{-17} \]
\[ 0.5\phi = 0.39 \times 10^{-17} \]
\[ \phi = 0.78 \times 10^{-17} \, \text{J} \]

4. **Calculate the energy of the photon with a wavelength of 3.846 nm.**
First, convert the wavelength to meters:
\[ \lambda = 3.846 \, \text{nm} = 3.846 \times 10^{-9} \, \text{m} \]
The energy of the photon can also be calculated using:
\[ E = \frac{h \cdot c}{\lambda} \]
where \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)).
\[ E = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{3.846 \times 10^{-9}} \]
\[ E \approx 5.17 \times 10^{-16} \, \text{J} \]

5. **Calculate the kinetic energy of the emitted photoelectron.**
Using the work function calculated earlier:
\[ KE = E - \phi \]
\[ KE = 5.17 \times 10^{-16} - 0.78 \times 10^{-17} \]
\[ KE \approx 4.39 \times 10^{-16} \, \text{J} \]

### Final Answer:
The energy of the photoelectron emitted when the metal is irradiated with light of wavelength 3.846 nm is approximately \( 4.39 \times 10^{-16} \, \text{J} \).