Light of frequency `7.21xx10^(14)Hz` is incident on a metal surface. Electrons with a maximum speed of `6.0xx10^(5)ms^(-1)` are ejected from the surface. What is the threshold frequency for photoemission of electrons? `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`.

Options

A. `2.32xx10^(14)Hz`
B. `2.32xx10^(12)Hz`
C. `4.74xx10^(14)Hz`
D. `4.74xx10^(12)Hz`

Explanation

To find the threshold frequency for the photoemission of electrons, we can use Einstein's photoelectric equation, which relates the kinetic energy of the emitted electrons to the energy of the incident photons and the work function of the metal.

### Step-by-Step Solution

1. **Identify Given Values:**
- Frequency of incident light, \( \nu = 7.21 \times 10^{14} \, \text{Hz} \)
- Maximum speed of ejected electrons, \( v = 6.0 \times 10^{5} \, \text{m/s} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of electron, \( m_e = 9.1 \times 10^{-31} \, \text{kg} \)

2. **Calculate the Kinetic Energy of the Ejected Electrons:**
The kinetic energy (KE) of the electrons can be calculated using the formula:
\[ KE = \frac{1}{2} m_e v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 9.1 \times 10^{-31} \, \text{kg} \times (6.0 \times 10^{5} \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 9.1 \times 10^{-31} \times 3.6 \times 10^{11} \]
\[ KE = 1.63 \times 10^{-19} \, \text{J} \]

3. **Use Einstein's Photoelectric Equation:**
According to Einstein's photoelectric equation:
\[ KE = h\nu - h\nu_0 \]
Rearranging for the threshold frequency \( \nu_0 \):
\[ h\nu_0 = h\nu - KE \]
\[ \nu_0 = \nu - \frac{KE}{h} \]

4. **Substituting Values:**
Substituting the values into the equation:
\[ \nu_0 = 7.21 \times 10^{14} - \frac{1.63 \times 10^{-19}}{6.63 \times 10^{-34}} \]
Calculating \( \frac{1.63 \times 10^{-19}}{6.63 \times 10^{-34}} \):
\[ \frac{1.63 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 2.46 \times 10^{14} \]
Now substituting back:
\[ \nu_0 = 7.21 \times 10^{14} - 2.46 \times 10^{14} \]
\[ \nu_0 \approx 4.75 \times 10^{14} \, \text{Hz} \]

5. **Final Result:**
The threshold frequency for the photoemission of electrons is approximately:
\[ \nu_0 \approx 4.74 \times 10^{14} \, \text{Hz} \]