The work functions of metals `A` and `B` are in the ratio `1 : 2`. If light of frequencies `f` and `2 f` are incident on the surfaces of `A` and `B` respectively , the ratio of the maximum kinetic energy of photoelectrons emitted is ( `f` is greater than threshold frequency of `A , 2f` is greater than threshold frequency of `B`)

Options

A. `1 : 1`
B. `1 : 2`
C. `1 : 3`
D. `1 : 4`

Answer

To solve the problem, we will use Einstein's photoelectric equation, which states that the maximum kinetic energy (K.E.) of the emitted photoelectrons is given by:

\[ K.E. = h\nu - \phi \]

where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

### Step 1: Define the work functions
Let the work function of metal A be \( \phi_A \) and that of metal B be \( \phi_B \). According to the problem, the ratio of the work functions is given as:

\[ \frac{\phi_A}{\phi_B} = \frac{1}{2} \]

From this, we can express \( \phi_B \) in terms of \( \phi_A \):

\[ \phi_B = 2\phi_A \]

### Step 2: Write the kinetic energy equations for both metals
For metal A, the frequency of incident light is \( f \):

\[ K.E._A = h f - \phi_A \]

For metal B, the frequency of incident light is \( 2f \):

\[ K.E._B = h (2f) - \phi_B \]

Substituting \( \phi_B \) from the previous step:

\[ K.E._B = h (2f) - 2\phi_A \]

### Step 3: Substitute and simplify the equations
Now we have:

1. \( K.E._A = h f - \phi_A \)
2. \( K.E._B = 2h f - 2\phi_A \)

### Step 4: Find the ratio of maximum kinetic energies
We need to find the ratio \( \frac{K.E._A}{K.E._B} \):

\[ \frac{K.E._A}{K.E._B} = \frac{h f - \phi_A}{2h f - 2\phi_A} \]

Factoring out common terms in the denominator:

\[ = \frac{h f - \phi_A}{2(h f - \phi_A)} = \frac{1}{2} \]

### Conclusion
Thus, the ratio of the maximum kinetic energy of photoelectrons emitted from metals A and B is:

\[ \frac{K.E._A}{K.E._B} = 1 : 2 \]