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Step-by-Step Solution
Step 1: Identify the given photon energies and work function
The problem provides two different photon energies:
Photon Energy 1, $E_1 = 1\text{ eV}$
Photon Energy 2, $E_2 = 2.5\text{ eV}$
and the metal surface has a work function,
$W = 0.5\text{ eV}$.
Step 2: Write the formula for maximum kinetic energy
According to Einstein's photoelectric equation:
$$K.E._\text{max} = E - W,$$
where
$E$ is the incident photon energy and
$W$ is the work function of the metal.
Step 3: Calculate the maximum kinetic energy for each photon
1) For the photon of energy $E_1 = 1\text{ eV}$:
$$K.E._{1,\text{max}} = 1\text{ eV} - 0.5\text{ eV} = 0.5\text{ eV}.$$
2) For the photon of energy $E_2 = 2.5\text{ eV}$:
$$K.E._{2,\text{max}} = 2.5\text{ eV} - 0.5\text{ eV} = 2.0\text{ eV}.$$
Step 4: Relate kinetic energy to electron velocity
The maximum kinetic energy can be expressed as:
$$K.E._\text{max} = \frac{1}{2} m v^2.$$
Thus for each photon, we have
$$\frac{1}{2} m v_1^2 = 0.5\text{ eV}, \quad \frac{1}{2} m v_2^2 = 2.0\text{ eV}.$$
Step 5: Determine the ratio of the electron speeds
From the above equations, the ratio of velocities $v_1$ and $v_2$ is:
$$
\frac{v_1}{v_2}
= \sqrt{\frac{K.E._{1,\text{max}}}{K.E._{2,\text{max}}}}
= \sqrt{\frac{0.5\text{ eV}}{2.0\text{ eV}}}
= \sqrt{\frac{1}{4}}
= \frac{1}{2}.
$$
Hence the ratio $v_1 : v_2$ is $1 : 2$.
Final Answer
The ratio of the maximum speeds of the emitted electrons is 1 : 2.
