In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will be

Options

A. `v(3/4)^(1//2)`
B. `v(3/4)^(1//2)`
C. less than `v(4/3)^(1//2)`
D. greater than `v(4/3)^(1//2)`

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the maximum kinetic energy of emitted electrons to the energy of the incoming photons. The equation is given by:

\[ K.E. = \frac{hc}{\lambda} - \phi_0 \]

where:
- \( K.E. \) is the maximum kinetic energy of the emitted electron,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi_0 \) is the work function of the material.

### Step 1: Write the equation for the initial wavelength \( \lambda \)

For the initial wavelength \( \lambda \), the maximum kinetic energy of the emitted electron can be expressed as:

\[ \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \phi_0 \tag{1} \]

### Step 2: Write the equation for the new wavelength \( \frac{3\lambda}{4} \)

When the wavelength is changed to \( \frac{3\lambda}{4} \), the maximum kinetic energy of the emitted electron becomes:

\[ \frac{1}{2} m v'^2 = \frac{hc}{\frac{3\lambda}{4}} - \phi_0 \tag{2} \]

### Step 3: Simplify the equation for the new wavelength

Substituting \( \frac{hc}{\frac{3\lambda}{4}} \) into equation (2):

\[ \frac{1}{2} m v'^2 = \frac{4hc}{3\lambda} - \phi_0 \]

### Step 4: Divide equation (2) by equation (1)

Now, we will divide equation (2) by equation (1):

\[ \frac{\frac{1}{2} m v'^2}{\frac{1}{2} m v^2} = \frac{\frac{4hc}{3\lambda} - \phi_0}{\frac{hc}{\lambda} - \phi_0} \]

This simplifies to:

\[ \frac{v'^2}{v^2} = \frac{\frac{4hc}{3\lambda} - \phi_0}{\frac{hc}{\lambda} - \phi_0} \]

### Step 5: Simplify the right-hand side

To simplify the right-hand side, we can express it as follows:

\[ \frac{v'^2}{v^2} = \frac{4hc - 3\lambda \phi_0}{3hc - 3\lambda \phi_0} \]

### Step 6: Analyze the expression

Since \( \phi_0 \) is a positive quantity, we can conclude that \( \frac{4hc - 3\lambda \phi_0}{3hc - 3\lambda \phi_0} \) is greater than \( \frac{4}{3} \) because the numerator is larger than the denominator.

Thus, we have:

\[ \frac{v'^2}{v^2} > \frac{4}{3} \]

### Step 7: Take the square root

Taking the square root of both sides gives:

\[ \frac{v'}{v} > \sqrt{\frac{4}{3}} \implies v' > v \cdot \sqrt{\frac{4}{3}} \]

### Conclusion

Therefore, the speed of the fastest electron when the wavelength is changed to \( \frac{3\lambda}{4} \) will be greater than \( v \cdot \sqrt{\frac{4}{3}} \).

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