If the work function of a metal is 6.63 eV, then find its threshold frequency for photoelectric effect.

The correct option is 1.6 × 10¹⁵ Hz.

Step-by-Step Explanation:

1. Understand the Formula:
The work function ($\Phi$) of a metal is the minimum energy required to eject an electron from its surface. It is related to the threshold frequency (${\nu }_0$) by the equation:
$\Phi =h{\nu }_0$
where:
• $h$ is Planck's constant, which is approximately $6.63\times {10}^{-34}\text{J·s}$ (or $\text{kg·m}{2}^{}\text{/s}$).
• ${\nu }_0$ is the threshold frequency in Hertz (Hz).

2. Convert Work Function to Joules:
The given work function is in electron-volts (eV):
$\Phi =6.63\text{eV}$
To convert this energy into Joules (J), we use the conversion factor $1\text{eV}=1.6\times {10}^{-19}\text{J}$:
$\Phi =6.63\times 1.6\times {\mathrm{10}}^{-19}\text{J}$

3. Calculate the Threshold Frequency:
Rearranging the work function formula to solve for threshold frequency (${\nu }_0$):
${\nu }_0=\frac{\Phi }{h}$
Substitute the values into the equation:
${\nu }_0=\frac{6.63\times 1.6\times {\mathrm{10}}^{-19}\text{J}}{6.63\times {\mathrm{10}}^{-34}\text{J·s}}$
Cancel the term $6.63$ from the numerator and denominator:
${\nu }_0=1.6\times \frac{{\mathrm{10}}^{-19}}{{\mathrm{10}}^{-34}}\text{Hz}$
Using exponent rules (${10}^a/{10}^b={10}^{a-b}$):
${\nu }_0=1.6\times {\mathrm{10}}^{-19-(-34)}\text{Hz}$
${\nu }_0=1.6\times {\mathrm{10}}^{15}\text{Hz}$

Thus, the threshold frequency for the photoelectric effect of the metal is 1.6 × 10¹⁵ Hz.