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Step-by-Step Solution
Step 1: Represent the known information using symbols
Let the initial energy of the incident radiation be $E$ (in electron-volts).
Since the energy is increased by 20%, the new energy becomes $1.2E$.
We are given:
• Initial kinetic energy of photoelectrons, $K_{1} = 0.5\,\text{eV}$
• Final kinetic energy of photoelectrons, $K_{2} = 0.8\,\text{eV}$
• Work function of the metal surface (to find), $\phi$.
Step 2: Write down the equations for photoelectric emission
According to the photoelectric equation:
$K_{\text{max}} = h \nu - \phi,$
where $h\nu$ is the incident photon energy and $\phi$ is the work function.
For the initial condition:
$K_{1} = E - \phi \quad \Rightarrow \quad 0.5 = E - \phi.$
For the final condition (with energy increased by 20%):
$K_{2} = 1.2E - \phi \quad \Rightarrow \quad 0.8 = 1.2E - \phi.$
Step 3: Solve the equations
From the first equation:
$ \phi = E - 0.5.$
From the second equation:
$ \phi = 1.2E - 0.8.$
Setting both expressions for $\phi$ equal:
$E - 0.5 = 1.2E - 0.8.$
Rearrange to solve for $E$:
$-0.5 + 0.8 = 1.2E - E,$
$0.3 = 0.2E,$
$E = \frac{0.3}{0.2} = 1.5\,\text{eV}.$
Step 4: Find the work function $ \phi$
Substitute $ E = 1.5\,\text{eV}$ into the expression for $\phi$:
$\phi = E - 0.5 = 1.5 - 0.5 = 1.0\,\text{eV}.$
Final Answer
The work function of the metal is $1.0\,\text{eV}$.
