If work function is 6.6 eV. The threshold frequency is x × 10¹⁴ Hz, Find x. (h = 6.6 × 10^–34 J.S)

The correct answer is 16.

Step-by-Step Explanation:

The work function ($\Phi$) of a metal is related to its threshold frequency (${\nu }_0$) by the photoelectric equation:
$\Phi =h{\nu }_0$

Where:
• $\Phi$ is the work function
• $h$ is Planck's constant
• ${\nu }_0$ is the threshold frequency

Given data:
• Work function, $\Phi =6.6\text{eV}$
• Planck's constant, $h=6.6\times {10}^{-34}\text{J·s}$
• Threshold frequency, ${\nu }_0=x\times {10}^{14}\text{Hz}$

First, we convert the work function from electron-volts (eV) to Joules (J). We know that:
$1\text{eV}=1.6\times {10}^{-19}\text{J}$

Therefore, the work function in Joules is:
$\Phi =6.6\times 1.6\times {10}^{-19}\text{J}$

Now, we substitute the values into the formula:
$6.6\times 1.6\times {10}^{-19}=(6.6\times {10}^{-34})\times {\nu }_0$

Solving for ${\nu }_0$:
${\nu }_0=\frac{6.6\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}$

We can cancel out the common factor of $6.6$ from the numerator and denominator:
${\nu }_0=1.6\times {10}^{-19-(-34)}$
${\nu }_0=1.6\times {10}^{15}\text{Hz}$

We need to express this in the form $x\times {10}^{14}\text{Hz}$:
${\nu }_0=16\times {10}^{14}\text{Hz}$

Comparing this with the given format $x\times {10}^{14}\text{Hz}$, we find:
$x=16$