Question

The threshold frequency of a metal is \(1.15 \times 10^{15}\) Hz. If electrons with kinetic energy of \(0.20\) eV are ejected when this metal surface is irradiated with photons of frequency '\(\nu\)', the value of \(\nu\) is (\(h=6.60 \times 10^{-34}\) Js, \(1\) eV \(=1.6 \times 10^{-19}\) J)

• A. \(1.20 \times 10^{14}\) Hz
• B. \(1.20 \times 10^{15}\) Hz
• C. \(1.98 \times 10^{14}\) Hz
• D. \(1.98 \times 10^{15}\) Hz

Hint:

For photoelectric-effect problems, first calculate the work function using \(\Phi=h\nu_0\), then apply Einstein's equation \(h\nu=\Phi+K_{\max}\).

Answer 1

The Correct Option is B

Concept: According to Einstein's photoelectric equation, \[ h\nu=\Phi+K_{\max} \] where \[ \Phi=h\nu_0 \] is the work function of the metal and \(\nu_0\) is the threshold frequency.

Step 1: Calculate the work function of the metal.
Given, \[ \nu_0=1.15\times10^{15}\ \text{Hz} \] \[ \Phi=h\nu_0 \] \[ \Phi=(6.60\times10^{-34})(1.15\times10^{15}) \] \[ \Phi=7.59\times10^{-19}\ \text{J} \]

Step 2: Convert kinetic energy into joules.
\[ K_{\max}=0.20\ \text{eV} \] Using \[ 1\ \text{eV}=1.6\times10^{-19}\ \text{J}, \] \[ K_{\max}=0.20\times1.6\times10^{-19} \] \[ K_{\max}=0.32\times10^{-19}\ \text{J} \] \[ K_{\max}=3.2\times10^{-20}\ \text{J} \]

Step 3: Calculate the frequency of the incident radiation.
Using \[ h\nu=\Phi+K_{\max}, \] \[ \nu=\frac{\Phi+K_{\max}}{h} \] \[ \nu= \frac{7.59\times10^{-19}+0.32\times10^{-19}} {6.60\times10^{-34}} \] \[ \nu= \frac{7.91\times10^{-19}} {6.60\times10^{-34}} \] \[ \nu\approx1.20\times10^{15}\ \text{Hz} \] \[ \boxed{1.20\times10^{15}\ \text{Hz}} \]