Question

The value of work function of a metal \(X\) is \(3.1\ \text{eV}\). The threshold frequency of it (in Hz) is
\[ (h=6.62\times 10^{-34}\ \text{Js}) \]

• A. \(6.49\times 10^{13}\)
• B. \(5.49\times 10^{13}\)
• C. \(6.49\times 10^{14}\)
• D. \(7.49\times 10^{14}\)

Hint:

Threshold frequency is the minimum frequency required for photoelectric emission: \[ \nu_0=\frac{\phi}{h} \] Always convert electron volt into joule before substitution.

Answer 1

The Correct Option is D

Step 1: Understand threshold frequency.
Threshold frequency $ \nu_0 $ is the minimum frequency of incident light that can eject an electron from the metal surface. Below this frequency, no photoelectrons are emitted regardless of intensity. It is related to work function by: \[ \phi = h\nu_0 \]
Step 2: Rearrange for threshold frequency.
\[ \nu_0 = \frac{\phi}{h} \] This tells us: greater the work function, higher the threshold frequency needed to overcome the surface barrier.
Step 3: Convert work function to joules.
Given $ \phi = 3.1\text{ eV} $ and $ 1\text{ eV} = 1.6\times10^{-19}\text{ J} $: \[ \phi = 3.1 \times 1.6\times10^{-19} = 4.96\times10^{-19}\text{ J} \]
Step 4: Calculate the threshold frequency.
With $ h = 6.62\times10^{-34}\text{ J s} $: \[ \nu_0 = \frac{4.96\times10^{-19}}{6.62\times10^{-34}} = \frac{4.96}{6.62}\times10^{15} \approx 0.749\times10^{15} = 7.49\times10^{14}\text{ Hz} \]
Step 5: Check the frequency range.
$ 7.49\times10^{14}\text{ Hz} $ lies in the ultraviolet range (just above visible light), which is physically reasonable for metals with work functions around 3 eV.
Step 6: Final answer.
\[ \boxed{\nu_0 = 7.49\times10^{14}\text{ Hz}} \]