Question:
The value of work function of a metal \(X\) is \(3.1\ \text{eV}\). The threshold frequency of it (in Hz) is
\[
(h=6.62\times 10^{-34}\ \text{Js})
\]
The value of work function of a metal \(X\) is \(3.1\ \text{eV}\). The threshold frequency of it (in Hz) is
\[ (h=6.62\times 10^{-34}\ \text{Js}) \]
\[ (h=6.62\times 10^{-34}\ \text{Js}) \]
Show Hint
Threshold frequency is the minimum frequency required for photoelectric emission:
\[
\nu_0=\frac{\phi}{h}
\]
Always convert electron volt into joule before substitution.
Updated On: Jun 25, 2026
- \(6.49\times 10^{13}\)
- \(5.49\times 10^{13}\)
- \(6.49\times 10^{14}\)
- \(7.49\times 10^{14}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use Einstein's photoelectric equation.
The threshold frequency is related to work function by \[ \phi=h\nu_0 \] where \[ \phi=\text{work function} \] \[ h=\text{Planck's constant} \] and \[ \nu_0=\text{threshold frequency} \] Therefore, \[ \nu_0=\frac{\phi}{h} \]
Step 2: Convert work function into joule.
Given: \[ \phi=3.1\ \text{eV} \] Since \[ 1\ \text{eV}=1.6\times 10^{-19}\ \text{J}, \] we get \[ \phi=3.1\times 1.6\times 10^{-19} \] \[ \phi=4.96\times 10^{-19}\ \text{J} \]
Step 3: Calculate threshold frequency.
Using \[ \nu_0=\frac{\phi}{h}, \] \[ \nu_0= \frac{4.96\times 10^{-19}} {6.62\times 10^{-34}} \] \[ \nu_0\approx 7.49\times 10^{14}\ \text{Hz} \]
Step 4: Final conclusion.
Hence, the threshold frequency is \[ \boxed{7.49\times 10^{14}\ \text{Hz}} \]
The threshold frequency is related to work function by \[ \phi=h\nu_0 \] where \[ \phi=\text{work function} \] \[ h=\text{Planck's constant} \] and \[ \nu_0=\text{threshold frequency} \] Therefore, \[ \nu_0=\frac{\phi}{h} \]
Step 2: Convert work function into joule.
Given: \[ \phi=3.1\ \text{eV} \] Since \[ 1\ \text{eV}=1.6\times 10^{-19}\ \text{J}, \] we get \[ \phi=3.1\times 1.6\times 10^{-19} \] \[ \phi=4.96\times 10^{-19}\ \text{J} \]
Step 3: Calculate threshold frequency.
Using \[ \nu_0=\frac{\phi}{h}, \] \[ \nu_0= \frac{4.96\times 10^{-19}} {6.62\times 10^{-34}} \] \[ \nu_0\approx 7.49\times 10^{14}\ \text{Hz} \]
Step 4: Final conclusion.
Hence, the threshold frequency is \[ \boxed{7.49\times 10^{14}\ \text{Hz}} \]
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