Light of wavelength `lamda` strikes a photoelectric surface and electrons are ejected with kinetic energy K. If K is to be increased to exactly twice its original value, the wavelength must be changed to `lamda'` such that

Options

A. `lamda'lt(lamda)/(2)`
B. `lamda,gt(lamda)/(2)`
C. `lamdagtlamda'gt(lamda)/(2)`
D. `lamda'=(lamda)/(2)`

Explanation

To solve the problem, we need to use the principles of the photoelectric effect, particularly Einstein's photoelectric equation, which is given by:

\[ E = W + K \]

where:
- \( E \) is the energy of the incident photon,
- \( W \) is the work function of the material (the minimum energy needed to eject an electron),
- \( K \) is the kinetic energy of the ejected electron.

The energy of a photon can also be expressed in terms of its wavelength \( \lambda \):

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

where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light.

### Step 1: Write the equation for the initial kinetic energy \( K \)

From the photoelectric equation, we can write:

\[ \frac{hc}{\lambda} = W + K \]

### Step 2: Write the equation for the new kinetic energy \( 2K \)

When the kinetic energy is doubled, we have:

\[ \frac{hc}{\lambda'} = W + 2K \]

### Step 3: Set up the equations

We now have two equations:

1. \( \frac{hc}{\lambda} = W + K \) (1)
2. \( \frac{hc}{\lambda'} = W + 2K \) (2)

### Step 4: Rearrange both equations to isolate \( W \)

From equation (1):

\[ W = \frac{hc}{\lambda} - K \]

From equation (2):

\[ W = \frac{hc}{\lambda'} - 2K \]

### Step 5: Set the two expressions for \( W \) equal to each other

Equating both expressions for \( W \):

\[ \frac{hc}{\lambda} - K = \frac{hc}{\lambda'} - 2K \]

### Step 6: Rearrange the equation

Rearranging gives:

\[ \frac{hc}{\lambda} + K = \frac{hc}{\lambda'} \]

### Step 7: Solve for \( \lambda' \)

Now, we can express \( \lambda' \) in terms of \( \lambda \):

\[ \frac{hc}{\lambda'} = \frac{hc}{\lambda} + K \]

Taking the reciprocal gives:

\[ \lambda' = \frac{hc}{\frac{hc}{\lambda} + K} \]

### Step 8: Simplify the expression

To simplify, we multiply both sides by \( \lambda \):

\[ \lambda' = \frac{hc \lambda}{hc + \lambda K} \]

### Step 9: Analyze the relationship between \( \lambda \) and \( \lambda' \)

Since \( K \) is positive, we can see that \( \lambda' \) will be less than \( \lambda \) because the denominator \( hc + \lambda K \) is greater than \( hc \). Thus, we conclude:

\[ \lambda' < \lambda \]

### Final Result

The wavelength \( \lambda' \) must be less than the original wavelength \( \lambda \) for the kinetic energy to be doubled.