Question

A light of frequency \( x \) Hz when falls on a metal plate emits electrons that have double the kinetic energy compared to when light of frequency \( y \) Hz falls on the same plate. The threshold frequency of the metal in Hz is:

• A. \( y - x \)
• B. \( x - y \)
• C. \( x - 2y \)
• D. \( 2y - x \)

Hint:

Always isolate the threshold frequency \( \nu_0 \) when equations involve multiple frequencies and their effects on electron emission.

Answer 1

The Correct Option is D

1. Understand the Photoelectric Effect

When light of a suitable frequency illuminates a metal surface, electrons are emitted. The maximum kinetic energy of these electrons depends on the frequency of the light and the work function of the metal.

2. Einstein's Photoelectric Equation

Einstein's photoelectric equation relates the energy of the incident photon (\( E \)), the work function of the metal (\( \Phi \)), and the maximum kinetic energy of the emitted electron (\( K_{\text{max}} \)):

\[ E = \Phi + K_{\text{max}} \]

where:

• \( E = hf \) (\( h \) is Planck's constant, and \( f \) is the frequency of light)
• \( \Phi \) is the work function
• \( K_{\text{max}} \) is the maximum kinetic energy

3. Apply the Equation to Both Cases

For light of frequency \( x \) Hz: \( hx = \Phi + K_{\text{max}_1} \)

For light of frequency \( y \) Hz: \( hy = \Phi + K_{\text{max}_2} \)

4. Apply the Given Condition

We are given that \( K_{\text{max}_1} = 2K_{\text{max}_2} \).

5. Substitute and Solve

Substitute \( K_{\text{max}_1} \) into the first equation:

\[ hx = \Phi + 2K_{\text{max}_2} \]

Subtract the second equation from this:

\[ hx - hy = \Phi + 2K_{\text{max}_2} - \Phi - K_{\text{max}_2} \]

Simplify:

\[ h(x - y) = K_{\text{max}_2} \]

Substitute \( K_{\text{max}_2} \) from the second equation:

\[ h(x - y) = hy - \Phi \]

Solve for \( \Phi \):

\[ \Phi = hy - h(x - y) = 2hy - hx \] 6. Find the Threshold Frequency

The threshold frequency (\( f_{\text{threshold}} \)) is the minimum frequency required for photoemission. At this frequency, \( K_{\text{max}} = 0 \).

Substitute \( K_{\text{max}} = 0 \) and \( \Phi = 2hy - hx \) into the second equation:

\[ hf_{\text{threshold}} = 2hy - hx + 0 \] \[ f_{\text{threshold}} = 2y - x \] Therefore, the threshold frequency of the metal is \( 2y - x \) Hz.

The correct answer is (4) \( 2y - x \).

Answer 2

To solve the problem, we need to find the threshold frequency of a metal when different light frequencies are used to emit electrons from the metal plate.

1. Understanding the Photoelectric Effect:
The photoelectric effect is explained by Einstein’s equation for the energy of emitted electrons:

\[ E_k = h \cdot f - \phi \] where:
\( E_k \) is the kinetic energy of the emitted electrons,
\( h \) is Planck's constant,
\( f \) is the frequency of the light falling on the metal plate,
\( \phi \) is the threshold frequency of the metal.

2. Relating the Given Information:
We are given that when light of frequency \( x \) Hz falls on the metal plate, the emitted electrons have double the kinetic energy compared to the kinetic energy of electrons emitted when light of frequency \( y \) Hz falls on the same plate.

Using the equation for the photoelectric effect for both frequencies:
For \( x \) Hz, the kinetic energy is: \[ E_k(x) = h \cdot x - \phi \]
For \( y \) Hz, the kinetic energy is: \[ E_k(y) = h \cdot y - \phi \]

We are told that the kinetic energy for \( x \) Hz is double the kinetic energy for \( y \) Hz: \[ h \cdot x - \phi = 2 \cdot (h \cdot y - \phi) \]

3. Solving the Equation:
Simplifying the equation:

\[ h \cdot x - \phi = 2h \cdot y - 2\phi \] \[ h \cdot x - 2h \cdot y = -\phi \] \[ h \cdot (x - 2y) = -\phi \] Thus, the threshold frequency \( \phi \) is given by: \[ \phi = h \cdot (2y - x) \] Therefore, the threshold frequency of the metal is \( 2y - x \).

4. Conclusion:
The threshold frequency of the metal is \( 2y - x \) Hz.

Final Answer:
The correct answer is (D) \( 2y - x \).