Photoelectric Effect

To solve this problem, let's analyze each statement individually and use our understanding of the physics of photons, specifically their momentum and energy properties.

1. Statement I: "Two photons having equal linear momenta have equal wavelengths."
   - The momentum of a photon ( ) is given by the equation , where is Planck's constant and is the wavelength of the photon.
   - If two photons have equal linear momentum, then: - Simplifying this equation yields , indicating the wavelengths are equal. Hence, Statement I is true.

1. Statement II: "If the wavelength of a photon is decreased, then the momentum and energy of a photon will also decrease."
   - As established, the momentum of a photon is inversely proportional to its wavelength: .
   - This means that if the wavelength ( ) decreases, the momentum ( ) actually increases, contrary to the statement's claim that it decreases. - Energy ( ) of a photon is given by (where is the speed of light).
   - As the wavelength decreases, the energy increases, again contrary to the statement which claims a decrease. Thus, Statement II is false.

Combining these analyses, we conclude that the correct answer is: Statement I is true but Statement II is false.

The correct answer is (D) : 3.42 eV
Frequency of EM waves

and

Energy of one photon of these waves

and

3.95 eV and 5.93 eV
⇒ Energy of maximum energetic electrons
= 5.93 – 2.50
= 3.43 eV

To solve the problem, we need to evaluate the resultant intensities at points A and B where two light beams interfere. The given intensities are 4I and 9I; these represent the intensities of the individual light beams.

Step 1: Calculate Resultant Intensity at Point A
At point A, the phase difference ϕ = 0, meaning the waves are in phase. The formula for resultant intensity I_res when two coherent beams interfere is: I_res = I₁ + I₂ + 2√(I₁ I₂) cos ϕ.
Substituting I₁ = 4I, I₂ = 9I, and cos(ϕ) = cos(0) = 1:
I_res,A = 4I + 9I + 2√(4I * 9I)
= 13I + 12I = 25I.

Step 2: Calculate Resultant Intensity at Point B
At point B, the phase difference ϕ = π, meaning the waves are out of phase. Use the same formula but with cos(ϕ) = cos(π) = -1:
I_res,B = 4I + 9I + 2√(4I * 9I) * (-1)
= 13I - 12I = 1I.

Step 3: Determine the Difference of Resultant Intensities
The difference between the intensities at points A and B is:
ΔI = I_res,A - I_res,B = 25I - 1I = 24I.

The computed intensity difference ΔI = 24I fits within the expected range of 24,24. Hence, the difference of resultant intensities, at points A and B, is 24I.

The correct answer is option (C) : 2 eV

On extrapolating the graph , the graph cuts the yaxis at -2

at v=0,v_s=-2v

The photoelectric effect is described by Einstein's photoelectric equation: where:
is the energy of the incident photon,
is the work function (minimum energy required to emit an electron),
is the maximum kinetic energy of the emitted photoelectrons.
Analysis of statements:
True. The photocurrent (number of photoelectrons emitted per second) is proportional to the intensity of the incident radiation, which determines the number of photons.
False. The maximum kinetic energy of photoelectrons depends on the frequency, not the intensity.
True. The maximum kinetic energy depends on the frequency of the incident light ( ) through the equation .
False. The emission of photoelectrons depends on the frequency of the light exceeding the threshold frequency, not intensity.
False. The maximum kinetic energy depends on the frequency.
Final Answer: The correct statements are:

Statement I: Ultraviolet (UV) rays have the highest frequency among the given options (microwaves, infrared rays, and ultraviolet rays). Since the energy of a photon is directly proportional to its frequency ( ), UV rays are the most effective for the emission of electrons from a metallic surface. Hence, Statement I is true.

Statement II: The maximum kinetic energy of photoelectrons is given by the photoelectric equation: where is Planck's constant, is the frequency of the incident light, and is the threshold frequency. From the equation, it is clear that is directly proportional to , not inversely proportional.

Hence, Statement II is false. Thus, the correct answer is .

1. Threshold Wavelength:
   • Photoelectric emission occurs if .
   • Given .
2. Analysis of Radiation:
   • Infra-red radiation has (no emission).
   • Ultra-violet radiation has (emission occurs).

Options that are correct: C and D

The problem involves the photoelectric effect, where ultraviolet (UV) light of energy is incident on a metal surface, causing electrons to be ejected. To find the maximum kinetic energy of these photoelectrons, we use the photoelectric equation given by Albert Einstein:

Where:

• is the maximum kinetic energy of the photoelectrons.
• is the energy of the incident photon.
• is the work function of the metal, the minimum energy required to eject an electron from the surface of the metal.

Given:

• Energy of the incident UV light,
• Work function of the metal,

Substituting these values into the equation:

Thus, the maximum kinetic energy of the ejected photoelectrons is 1 eV.

Why other options are incorrect:

• is the energy of the incident light, not the kinetic energy.
• is the work function, not the kinetic energy.
• is obtained by incorrectly adding the incident light energy and the work function instead of subtracting.

The correct answer is indeed 1 eV.

Understanding the Photoelectric Effect and Current-Voltage Characteristics:
In the photoelectric effect, the photoelectric current depends on the intensity of the incident light and the applied potential difference .

The key points to note are:
- For a given frequency (or wavelength), the stopping potential remains constant, as it depends only on the frequency of the incident light and not on its intensity.
- The saturation current (maximum current) is proportional to the intensity of the incident light. Therefore, higher intensity light (e.g., ) results in a higher saturation current than lower intensity light (e.g., ).

Selecting the Correct Graph:
Since the wavelength (or frequency) of the light is the same for both intensities, the stopping potential will be the same for both and . However, since , the saturation current for will be greater than that for .
Option (3) correctly shows:
- The stopping potential is the same for both intensities.
- The saturation current for is greater than for , consistent with the higher intensity of .

Conclusion:
Therefore, the correct graph is Option (3), as it accurately represents the photoelectric current variation with applied potential for two different light intensities of the same wavelength.

Given:

The problem involves the relation between the incident frequency and the threshold frequency for the emission of photoelectrons. We are given the following inequality: where is the incident frequency, and is the threshold frequency.

Step 1: Understanding the relationship between incident frequency and threshold frequency

The inequality implies that the incident frequency is less than twice the threshold frequency . In this context, it indicates that the incident light has a frequency that is less than the required threshold frequency needed for the emission of photoelectrons.

Step 2: Explanation of the photoelectric effect

In the photoelectric effect, photoelectrons are emitted from the surface of a material when it is illuminated by light of a frequency greater than or equal to the threshold frequency . If the incident frequency is less than the threshold frequency, the energy of the photons is insufficient to release the photoelectrons.

Step 3: Conclusion from the given condition

Since is less than the threshold frequency , the photons do not possess enough energy to overcome the work function of the material. This leads to the non-emission of photoelectrons.

Conclusion:

Therefore, the current, which depends on the number of photoelectrons emitted, is zero:

Final Answer:

No emission of photoelectrons occurs, leading to no current being generated in this case.

To find the threshold frequency for the photoelectric effect, we can use the relationship between the work function ( ) and the threshold frequency ( ). The formula that relates these quantities is based on Einstein's photoelectric equation:

where:

• is the work function of the metal, given in electron volts (eV).
• is Planck's constant, which is approximately .
• is the threshold frequency we need to find.

Given:

Plugging in these values into the equation:

Solving for :

Calculating the division gives:

Rounding off, this corresponds to .

Therefore, the threshold frequency for the photoelectric effect is 1.6 × 10¹⁵ Hz.

The stopping potential ( ) is related to frequency ( ) by the equation:

The slope of the graph gives , confirming Statement-I. However, has a higher work function, meaning that for the same incident frequency, the kinetic energy of emitted photoelectrons will be lower. Therefore, Statement-II is incorrect.

To determine the correct expression for the de-Broglie wavelength of the emitted electrons from a photo-emissive substance, we need to consider the photoelectric effect and apply the concept of energy conservation.

The energy of the incident photon can be given by:

where is the wavelength of the incident radiation, is the Planck's constant, and is the speed of light.

According to the photoelectric effect, the energy of the incident photon is used to overcome the work function of the substance and to provide kinetic energy to the emitted electron. Therefore, we have:

where is the kinetic energy of the emitted electron with velocity and mass .

The longest wavelength of radiation that can emit photoelectron is , which corresponds to the work function , so:

Substitute into the equation for the incident energy:

Simplify to find the expression for the kinetic energy:

Now, the de-Broglie wavelength of the electron is given by:

To find , use the equation for kinetic energy:

Solving for , we have:

Replacing in the de-Broglie wavelength formula gives us:

Thus, the correct option is: .

The energy of the incident photon is: For the photoelectric effect to occur, the energy of the photon must be greater than the work function of the metal.

For Cs (work function = 1.9 eV):

Since , photoelectric effect is possible for Cs.

For Li (work function = 2.5 eV):

Since , photoelectric effect is not possible for Li.

Thus, the answer is .

To determine the color of light required to cause the emission of photoelectrons from a metal with a work function of 3 eV, we need to understand the concept of the photoelectric effect. In this phenomenon, electrons are emitted from a metal surface when light of sufficient energy strikes it.

The energy required to eject an electron is given by the equation:

Where:

• is the energy of the incident light (must be equal to or greater than the work function),
• is Planck's constant , and
• is the frequency of the light.

The work function of the metal is the minimum energy required to remove an electron from the metal surface. For this problem:

Since energy also relates to wavelength via:

Where:

• is the speed of light , and
• is the wavelength.

Given , we solve for the wavelength :

Converting into nanometers:

This wavelength corresponds to blue light in the visible spectrum, which ranges from about 450 nm to 495 nm. Thus, blue light is indeed adequate to cause the photoelectric emission for this metal.

Hence, the correct answer is: Blue.

Analysis of Assertion (A) and Reason (R):

Assertion (A): The statement claims that the emission of electrons in the photoelectric effect can be inhibited by applying a sufficiently negative electric potential to the photoemissive material. This is true because applying a negative potential, known as the stopping or retarding potential, counteracts the forward motion of the photo-emitted electrons, preventing them from reaching the anode.

Reason (R): The reason states that the negative electric potential required to stop the emission of electrons from the surface of a photoemissive substance varies linearly with the frequency of incident radiation. This is also correct. According to the photoelectric effect, the stopping potential is related to the frequency of the incident radiation by the equation , where is the electron charge, is Planck's constant, and is the work function of the material. This equation indeed shows a linear relationship between the stopping potential and frequency.

Thus, both Assertion (A) and Reason (R) are true. However, while (R) is true and states a concept related to the photoelectric effect, it does not directly explain why applying a negative potential suppresses electron emission described in (A). The suppression is primarily due to the opposing electric field, not specifically because of the linear relationship with frequency.

Conclusion:

The correct answer is: Both (A) and (R) are true but (R) is not the correct explanation of (A).

The problem given is based on the photoelectric effect, a phenomenon where light incident on a metal surface causes the emission of electrons. We are provided with the work function of the metal and the stopping potential, both of which are essential in determining the wavelength of the incident electromagnetic wave.

Given:

• Work function ( ) = 2.14 eV
• Stopping potential ( ) = 2 V
• Product of Planck's constant and speed of light ( ) = 1242 eV·nm

To find the wavelength ( ) of the electromagnetic wave, we need to follow these steps:

The total energy ( ) of the incident photon is given by the sum of the work function and the energy required to overcome the stopping potential:

The energy ( ) of the incident photon can also be expressed in terms of its wavelength ( ) using the relation:

Substitute the known values and solve for :

Rearrange the equation to solve for :

Thus, the wavelength of the electromagnetic wave is 300 nm.

Let's verify the options given:

• 400 nm
• 600 nm
• 200 nm
• 300 nm

The correct answer from the options is indeed 300 nm.

Assertion (A): In photoelectric effect, on increasing the intensity of incident light, the stopping potential increases.

Reason (R): Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than the threshold frequency.

Concept Used:

The photoelectric effect is governed by Einstein’s photoelectric equation:

where is the maximum kinetic energy of emitted photoelectrons, is Planck’s constant, is the frequency of the incident light, and is the work function of the metal. The stopping potential is related to as:

Step-by-Step Solution:

Step 1: Analyze the effect of intensity on stopping potential.

From the equation , we see that stopping potential depends only on the frequency of the incident light, not on its intensity. Therefore, increasing the intensity of light does not increase the stopping potential. The stopping potential remains the same if the frequency is unchanged.

Step 2: Analyze the Reason (R).

When the intensity of light increases, the number of photons incident per second on the metal surface increases. If each photon has energy greater than the work function ( ), more electrons are emitted per unit time. Hence, the rate of photoelectron emission increases with intensity.

Step 3: Evaluate the truth of A and R.

• Assertion (A) is false — stopping potential does not depend on intensity.
• Reason (R) is true — intensity affects the rate of emission, not the energy of photoelectrons.

Final Computation & Result:

Final Answer: Assertion (A) is false, but Reason (R) is true.

Concept:
From photoelectric effect, where is the work function.

For the same frequency (or wavelength), is constant. So, Thus, smaller work function ⇒ greater kinetic energy.

Graph Analysis:
In the graph of vs :

• The slope is same for all (since slope = )
• The x-intercept gives threshold frequency

So, the metal with smallest threshold frequency has smallest work function.

Conclusion:
The line which cuts the frequency axis at the leftmost point corresponds to the smallest . Hence, that metal gives maximum kinetic energy.

From the graph, this corresponds to:


Final Answer:

Step 1: The given electric field consists of two frequencies:
Step 2: In the photoelectric effect, only the highest frequency photon determines the maximum kinetic energy of emitted electrons.
Step 3: Energy of a photon: Using :
Step 4: Apply Einstein’s photoelectric equation: