When the electromagnetic radiations of frequencies `4 xx 10^(15) Hz` and `6 xx 10^(15) Hz` fall on the same metal, in different experiments, the ratio of maximum kinetic energy of electrons liberated is `1 : 3`. The threshold frequency for the metal is

Options

A. `2xx10^15` Hz
B. `1xx10^15` Hz
C. `3xx10^15` Hz
D. `1.67xx10^15` Hz

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between kinetic energy and frequency
The maximum kinetic energy (K.E.) of the emitted electrons can be expressed using the equation:
\[ K.E. = h f - h f_0 \]
where:
- \( K.E. \) is the maximum kinetic energy,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident radiation,
- \( f_0 \) is the threshold frequency of the metal.

### Step 2: Set up the equations for both frequencies
Let:
- \( f_1 = 4 \times 10^{15} \, \text{Hz} \)
- \( f_2 = 6 \times 10^{15} \, \text{Hz} \)

For frequency \( f_1 \):
\[ K.E._1 = h f_1 - h f_0 \]
For frequency \( f_2 \):
\[ K.E._2 = h f_2 - h f_0 \]

### Step 3: Use the given ratio of kinetic energies
We know that the ratio of maximum kinetic energies is given as:
\[ \frac{K.E._1}{K.E._2} = \frac{1}{3} \]

Substituting the expressions for \( K.E. \):
\[ \frac{h f_1 - h f_0}{h f_2 - h f_0} = \frac{1}{3} \]

### Step 4: Simplify the equation
Since \( h \) is a common factor, we can cancel it out:
\[ \frac{f_1 - f_0}{f_2 - f_0} = \frac{1}{3} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 3(f_1 - f_0) = f_2 - f_0 \]

### Step 6: Rearranging the equation
Expanding and rearranging the equation:
\[ 3f_1 - 3f_0 = f_2 - f_0 \]
\[ 3f_1 - f_2 = 2f_0 \]

### Step 7: Solve for the threshold frequency \( f_0 \)
Now, we can express \( f_0 \):
\[ f_0 = \frac{3f_1 - f_2}{2} \]

### Step 8: Substitute the values of \( f_1 \) and \( f_2 \)
Substituting the values:
\[ f_0 = \frac{3(4 \times 10^{15}) - (6 \times 10^{15})}{2} \]
\[ f_0 = \frac{12 \times 10^{15} - 6 \times 10^{15}}{2} \]
\[ f_0 = \frac{6 \times 10^{15}}{2} \]
\[ f_0 = 3 \times 10^{15} \, \text{Hz} \]

### Final Answer
The threshold frequency \( f_0 \) for the metal is:
\[ f_0 = 3 \times 10^{15} \, \text{Hz} \]

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