Photoelectric Effect

To determine the velocity of the emitted photoelectrons, we can apply the photoelectric effect formula:

Here, is the energy of the incident photons, is Planck's constant, is the frequency of the incident light, is the speed of light, and is the wavelength.

Given:

• Work function,
• Wavelength,
• The energy equivalent of

First, let's convert the work function from eV to joules:

Using Planck's constant, , and the speed of light, , we calculate the energy of the incident light:

Calculating the above expression:

This is the energy of the incident photon. The kinetic energy of the emitted photoelectron is calculated using the equation:

Substitute the known values:

Now solve for the velocity, , using the electron's mass :

Calculating the above expression:

Thus, the velocity of the emitted photoelectrons is approximately . Hence, the correct option is:

To solve this problem, we need to understand the relationship between the momentum and energy of protons and photons.

1. The kinetic energy ( ) of the proton is given as of its rest mass energy. The rest mass energy ( ) of a proton is given by Einstein’s mass-energy equivalence , where is the mass of the proton and is the speed of light.
2. Thus, the kinetic energy of the proton can be expressed as: .
3. For a proton, we know that the relationship between kinetic energy, momentum ( ), and mass-energy equivalence is: .
4. Setting the two expressions for kinetic energy equal, we have: .
   Solving for , we get: .
5. We simplify this to: .
6. A photon with the same kinetic energy as the proton will have energy .
7. For photons, energy ( ) is related to momentum ( ) by: .
   Since , we have: .
8. We are asked to find the ratio: .
   Substitute the values:
9. .
10. Solving further:
    \)
    .

Thus, the correct answer is .

The Fundamental Equation

Einstein's photoelectric equation establishes the relationship:

where:

• = Maximum kinetic energy of ejected electrons
• = Planck's constant ( Js)
• = Frequency of incident light
• = Work function (material-specific energy threshold)

The equation can be viewed as a linear function:

where:

• (Dependent variable)
• (Independent variable)
• (Slope)
• (Y-intercept)

Slope (h)

The positive slope indicates:

• Direct proportionality between kinetic energy and frequency
• Higher frequency light results in more energetic photoelectrons
• The slope's value equals Planck's constant

Y-intercept (-φ)

The negative intercept reveals:

• No electron emission occurs below the threshold frequency
• The magnitude represents the work function
• X-intercept ( ) marks the minimum frequency for emission

The correct answer is (D) :

Given:

• Work function (Φ) = 1 eV = 1.6 × 10⁻¹⁹ J
• Wavelength of light (λ) = 3000 Å = 3000 × 10⁻¹⁰ m = 3 × 10⁻⁷ m
• Planck's constant (h) = 6.63 × 10⁻³⁴ Js
• Speed of light (c) = 3 × 10⁸ m/s
• Electron mass (mₑ) = 9.11 × 10⁻³¹ kg

Step 1: Calculate photon energy (E)

Using Planck's equation:

Convert to eV:

Step 2: Determine kinetic energy of photoelectrons

Using Einstein's photoelectric equation:

Convert to joules:

Step 3: Calculate electron velocity

Using kinetic energy formula:

According to the photoelectric effect, the number of electrons emitted by a material depends on the intensity of the incident light, while the kinetic energy of the emitted electrons depends on the frequency of the incident light.
If both the intensity and frequency of the incident light are doubled, the following changes occur:
Doubling the intensity of the light: Increasing the intensity of light means more photons are incident on the material, leading to a higher number of electrons being emitted. Therefore, the saturation photocurrent will be doubled.
Doubling the frequency of the light: Increasing the frequency of light implies that each photon carries more energy. Consequently, the emitted electrons will have higher kinetic energy. However, doubling the frequency alone doesn't directly affect the saturation photocurrent.
In summary, doubling both the intensity and frequency of the incident light will result in the saturation photocurrent being doubled.
Therefore, the correct option is (A) the saturation photocurrent is doubled.

Energy of a photon is given by the equation:

Where:

• E = energy of the photon
• h = Planck's constant =
• c = speed of light =
• = wavelength of light

Convert the energy difference from eV to Joules:

For the initial wavelength of 500 nm:

The initial energy ( ) is calculated as:

For the new wavelength , the energy difference is given by:

Now, solving for :

Now, substitute the new energy into the equation:

Solving for :

\end{equation*}

Convert to nanometers:

Therefore, the approximate new wavelength is: 414 nm

After comparing with the answer choices: The correct option is , which is close to our calculated result.

Step 1: Recall Einstein’s photoelectric equation clearly:

The photoelectric equation is given by:

where:
= electronic charge,
= stopping potential,
= Planck’s constant,
= frequency of incident radiation,
= threshold frequency.

Step 2: Rewrite the equation in linear form ( ):

Expressing clearly in terms of frequency :

This equation is now in the standard linear form:

where:
,
,
(slope),
(y-intercept).

Step 3: Clearly identify slope and y-intercept from above comparison:

• Slope ( ):
• Y-intercept ( ):

Final Conclusion:
Slope: , Y-intercept: .

The energy of a photon is given by:
where
is Planck's constant,
is the speed of light, and
is the wavelength.

For photoemission to occur, the energy of the photon must be greater than or equal to the work function .
For , the photon energy is lower than the work function, meaning no photoemission occurs.

The photoelectric effect equation is:
where
is the energy of the incident light and
is the work function of the metal.

The kinetic energy of the photoelectrons is the difference between the energy of the incident photons and the energy required to overcome the work function.
For typical photoelectric effects, the kinetic energy is a fraction of the photon energy, and it is often observed to be less than but still significant.