The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal ( for which work function is `1.65 eV`) will be

Options

A. `4 xx 10^(10) Hz`
B. `4 xx 10^(11) Hz`
C. `4 xx 10^(14) Hz`
D. `4 xx 10^(-10) Hz`

Explanation

To find the lowest frequency of light that will cause the emission of photoelectrons from a metal surface with a given work function, we can use the photoelectric effect equation:

1. **Understand the relationship**: The energy of the incident photon must be equal to or greater than the work function (Φ) of the metal for photoemission to occur. This can be expressed as:
\[ E = h \nu \geq \Phi \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(\nu\) is the frequency of the light, and \(\Phi\) is the work function.

2. **Convert work function to Joules**: The work function is given as \(1.65 \, \text{eV}\). We need to convert this to Joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore,
\[ \Phi = 1.65 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.64 \times 10^{-19} \, \text{J} \]

3. **Use Planck's constant**: Planck's constant \(h\) is approximately:
\[ h = 6.63 \times 10^{-34} \, \text{J s} \]

4. **Rearranging the equation**: We need to find the frequency \(\nu\). Rearranging the equation gives:
\[ \nu = \frac{\Phi}{h} \]

5. **Substituting values**: Now substitute the values of \(\Phi\) and \(h\) into the equation:
\[ \nu = \frac{2.64 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J s}} \]

6. **Calculating the frequency**: Performing the calculation:
\[ \nu = 3.98 \times 10^{14} \, \text{Hz} \]

7. **Rounding off**: Rounding off gives us approximately:
\[ \nu \approx 4.0 \times 10^{14} \, \text{Hz} \]

Thus, the lowest frequency of light that will cause the emission of photoelectrons from the surface of the metal is approximately \(4.0 \times 10^{14} \, \text{Hz}\).