Find the difference of kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500Å and 5000Å. `h=6.62xx10^(-34)Js`.

Options

A. 1.61 eV
B. 2.47 eV
C. 3.96 eV
D. `3.96xx10^(-19)J`

Answer

To find the difference in kinetic energies of photoelectrons emitted from a surface by light of wavelengths 2500 Å and 5000 Å, we can follow these steps:

### Step 1: Understand the photoelectric effect
The kinetic energy (KE) of photoelectrons emitted from a surface is given by the equation:
\[ KE = \frac{hc}{\lambda} - \phi \]
where:
- \( KE \) is the kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the material.

### Step 2: Write the kinetic energy equations for both wavelengths
Let:
- \( \lambda_1 = 2500 \, \text{Å} = 2500 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)

The kinetic energies for the two wavelengths can be expressed as:
\[ KE_1 = \frac{hc}{\lambda_1} - \phi \]
\[ KE_2 = \frac{hc}{\lambda_2} - \phi \]

### Step 3: Find the difference in kinetic energies
To find the difference in kinetic energies, we subtract the two equations:
\[ KE_1 - KE_2 = \left(\frac{hc}{\lambda_1} - \phi\right) - \left(\frac{hc}{\lambda_2} - \phi\right) \]
The work function \( \phi \) cancels out:
\[ KE_1 - KE_2 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
\[ KE_1 - KE_2 = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) \]

### Step 4: Substitute the values
Now we can substitute the known values:
- \( h = 6.62 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^{8} \, \text{m/s} \)
- \( \lambda_1 = 2500 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 5000 \times 10^{-10} \, \text{m} \)

Calculating the difference:
\[ KE_1 - KE_2 = 6.62 \times 10^{-34} \times 3 \times 10^{8} \left(\frac{1}{2500 \times 10^{-10}} - \frac{1}{5000 \times 10^{-10}}\right) \]

### Step 5: Simplify the expression
Calculating the fractions:
\[ \frac{1}{2500 \times 10^{-10}} = \frac{1}{2.5 \times 10^{-7}} = 4 \times 10^{6} \]
\[ \frac{1}{5000 \times 10^{-10}} = \frac{1}{5 \times 10^{-7}} = 2 \times 10^{6} \]

Now substituting back:
\[ KE_1 - KE_2 = 6.62 \times 10^{-34} \times 3 \times 10^{8} \left(4 \times 10^{6} - 2 \times 10^{6}\right) \]
\[ KE_1 - KE_2 = 6.62 \times 10^{-34} \times 3 \times 10^{8} \times 2 \times 10^{6} \]

### Step 6: Calculate the final value
Calculating:
\[ KE_1 - KE_2 = 6.62 \times 10^{-34} \times 3 \times 10^{8} \times 2 \times 10^{6} \]
\[ = 6.62 \times 3 \times 2 \times 10^{-34 + 8 + 6} \]
\[ = 39.72 \times 10^{-20} \]
\[ = 3.972 \times 10^{-19} \, \text{J} \]

### Final Answer
The difference in kinetic energies of the photoelectrons emitted from the surface by light of wavelengths 2500 Å and 5000 Å is approximately:
\[ KE_1 - KE_2 \approx 3.96 \times 10^{-19} \, \text{J} \]

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