A physicist wishes to eject electrons by shining light on a metal surfac. The light source emits light of wavelenght of 450 nm. The table lists the only available metals and their work functions.
`{:("Metal",W_(0)(eV)),("Barium",2.5),("Lithium",2.3),("Tantalum",4.2),("Tungsten",4.5):}`
Which option correctly identifies the metal that will produce the most energetic electrons and their energies?

Options

A. Lithium, 0.45 eV
B. Tungsten 1.75 eV
C. Lithium, 2.30eV
D. Tungsten, 2.75 eV

Answer

To solve the problem, we need to determine which metal will produce the most energetic electrons when illuminated with light of wavelength 450 nm. We will follow these steps:

### Step 1: Calculate the energy of the incident light
The energy of the photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon (in eV),
- \(h\) is Planck's constant (\(4.1357 \times 10^{-15} \, \text{eV s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the light (in meters).

Given that the wavelength \(\lambda = 450 \, \text{nm} = 450 \times 10^{-9} \, \text{m}\), we can substitute the values into the formula.

### Step 2: Substitute the values into the formula
\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{450 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ E \approx 2.75 \, \text{eV} \]

### Step 3: Compare the energy with the work functions of the metals
Next, we will compare the calculated energy with the work functions of the available metals:
- Barium: \(W_0 = 2.5 \, \text{eV}\)
- Lithium: \(W_0 = 2.3 \, \text{eV}\)
- Tantalum: \(W_0 = 4.2 \, \text{eV}\)
- Tungsten: \(W_0 = 4.5 \, \text{eV}\)

### Step 4: Determine which metals can eject electrons
To eject electrons, the energy of the incident light must be greater than the work function of the metal. Therefore, we check:
- For Barium: \(2.75 \, \text{eV} > 2.5 \, \text{eV}\) (can eject)
- For Lithium: \(2.75 \, \text{eV} > 2.3 \, \text{eV}\) (can eject)
- For Tantalum: \(2.75 \, \text{eV} < 4.2 \, \text{eV}\) (cannot eject)
- For Tungsten: \(2.75 \, \text{eV} < 4.5 \, \text{eV}\) (cannot eject)

### Step 5: Calculate the kinetic energy of the ejected electrons
The kinetic energy (KE) of the ejected electrons can be calculated using the formula:
\[ KE = E - W_0 \]
- For Barium:
\[ KE = 2.75 \, \text{eV} - 2.5 \, \text{eV} = 0.25 \, \text{eV} \]
- For Lithium:
\[ KE = 2.75 \, \text{eV} - 2.3 \, \text{eV} = 0.45 \, \text{eV} \]

### Step 6: Identify the metal with the highest kinetic energy
Comparing the kinetic energies:
- Barium: \(0.25 \, \text{eV}\)
- Lithium: \(0.45 \, \text{eV}\)

Lithium produces the most energetic electrons.

### Final Answer
The metal that will produce the most energetic electrons is **Lithium** with a kinetic energy of **0.45 eV**.

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