The energy required to remove an electron from metal X is E = `3.31 xx 10^(-20)J`. Calculate the maximum wavelength of light that can be photoeject an electron from metal X :

Options

A. `4 mum`
B. `6 mum`
C. `7 mum`
D. `5 mum`

Answer

To solve the problem of calculating the maximum wavelength of light that can photoeject an electron from metal X, we will use the relationship between energy, wavelength, and the constants of physics. Here’s a step-by-step solution:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 2: Rearrange the formula to solve for wavelength
To find the wavelength, we can rearrange the formula:
\[ \lambda = \frac{hc}{E} \]

### Step 3: Substitute the known values
Now we will substitute the known values into the equation. We are given:
- \( E = 3.31 \times 10^{-20} \, \text{J} \)
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3.00 \times 10^8 \, \text{m/s} \)

Substituting these values into the equation:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{3.31 \times 10^{-20} \, \text{J}} \]

### Step 4: Calculate the wavelength
Now, we perform the calculation:
1. Calculate the numerator:
\[ 6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]
2. Now divide by the energy:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{3.31 \times 10^{-20}} \]
\[ \lambda \approx 6.00 \times 10^{-6} \, \text{m} \]

### Step 5: Convert to micrometers
To convert meters to micrometers (1 m = \( 10^6 \) µm):
\[ \lambda \approx 6.00 \times 10^{-6} \, \text{m} = 6.00 \, \mu m \]

### Final Answer
The maximum wavelength of light that can photoeject an electron from metal X is approximately:
\[ \lambda \approx 6.00 \, \mu m \]

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