A physicist wishes to eject electrons by shining light on a metal surface. The light source emits light of wavelength 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 metal(s) can be used to produce electrons by the photoelectric effect from given source of light?

Options

A. Barium only
B. Barium or lithium
C. Lithium, tantanlum ot tungsten
D. Tungsten or tantalum

Answer

To determine which metals can be used to produce electrons by the photoelectric effect 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 incident light can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \(E\) is the energy of the incident light,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3.00 \times 10^{8} \, \text{m/s}\)),
- \(\lambda\) 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:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^{8} \, \text{m/s})}{450 \times 10^{-9} \, \text{m}} \]

Calculating this gives:

\[ E \approx 4.95 \times 10^{-19} \, \text{J} \]

### Step 2: Convert energy from Joules to electron volts

To convert the energy from Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{C}\):

\[ E (\text{eV}) = \frac{E (\text{J})}{1.6 \times 10^{-19} \, \text{C}} \]

Substituting the energy we calculated:

\[ E (\text{eV}) = \frac{4.95 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \approx 3.09 \, \text{eV} \]

### Step 3: Compare the energy with the work functions of the metals

Now we will compare the calculated energy \(E \approx 3.09 \, \text{eV}\) 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 emit electrons

The photoelectric effect will occur if the energy of the incident light is greater than the work function of the metal.

- For Barium: \(3.09 \, \text{eV} > 2.5 \, \text{eV}\) (Yes, electrons can be ejected)
- For Lithium: \(3.09 \, \text{eV} > 2.3 \, \text{eV}\) (Yes, electrons can be ejected)
- For Tantalum: \(3.09 \, \text{eV} < 4.2 \, \text{eV}\) (No, electrons cannot be ejected)
- For Tungsten: \(3.09 \, \text{eV} < 4.5 \, \text{eV}\) (No, electrons cannot be ejected)

### Conclusion

The metals that can be used to produce electrons by the photoelectric effect with the given light source are **Barium and Lithium**.

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