Test: Dual Nature of Radiation & Matter - NEET MCQ

In the photoelectric effect, the work function is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal.Electrons ejected from a sodium metal surface were measured as an electric current.The minimum energy required to eject an electron from the surface is called the photoelectric work function.
This energy (work function) is a measure of how firmly a particular metal holds its electrons. The work function is important in applications involving electron emission from metals, as in photoelectric devices and cathode-ray tubes.

A photon is massless, has no electric charge, and is a stable particle. In vacuum, a photon has two possible polarization states. The photon is the gauge boson for electromagnetism, and therefore all other quantum numbers of the photon (such as lepton number, baryon number, and flavour quantum numbers) are zero.
 

Threshold frequency: - frequency of minimum energy required to remove electron
 
work function=3.3ev
f=3.3×1.6×10⁻¹⁹J​ /6.626×10⁻³⁴J−s
=0.8×10¹⁵Hz
=8×10¹⁴Hz

Solution:

• The energy of a photon can be calculated using the formula: E = hc / λ.
• Here, h is Planck's constant, which is 6.63 × 10^-34 J·s.
• The speed of light (c) is 3 × 10⁸ m/s.
• The wavelength (λ) is given as 5500 × 10^-10 m.

Substitute these values into the formula:

• E = [6.63 × 10^-34 J·s × 3 × 10⁸ m/s] / 5500 × 10^-10 m
• Calculating gives: E = [0.36 × 10^-26] / 10^-8
• This simplifies to: E = 3.6 × 10^-19 J

Convert energy from joules to electronvolts:

• 1 electronvolt (eV) = 1.6 × 10^-19 J
• Therefore, E = 3.6 × 10^-19 J / 1.6 × 10^-19 J/eV
• The result is approximately 2.26 eV.

The forces acting on an electron in uniform electric and magnetic fields pointing in the same direction are explained below:

• Magnetic Force: Since the electron is moving parallel to the fields, the magnetic force (v × B) is zero.
• Electric Force: The electric field exerts a force (F_e = -E) on the electron, acting opposite to its direction of motion.

As a result:

• The electron experiences a backward force due to the electric field.
• This force causes the electron to decelerate, reducing its velocity.

The correct answer would be option B. Applying a very strong magnetic field.
Applying a very strong magnetic field to a metal we don’t know the electron emission.

A photon of wavelength 6000 nm collides with an electron at rest. After scattering, the wavelength of the scattered photon is found to change by exactly one Compton wavelength.

Frequency=C/ λ
λ=h/P
frequency=C P/ h
f=(3×10⁸×3.3×10^-29)/6.6×10^-34
f=3× 10¹³/2
f=1.5×10¹³ Hz

Φ= hνλ => 2eV= 6.626 x 10^-34 x ν

2 x 1.6 x 10^-19= 6.626 x 10^-34 x ν

On solving we get ν = 4.6 x 10¹⁴ Hz.

Photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it. In a broader definition, the radiant energy may be infrared, visible, or ultraviolet light, X-rays, or gamma rays; the material may be a solid, liquid, or gas; and the released particles may be ions (electrically charged atoms or molecules) as well as electrons.

In a photon-electron collision both total energy and total momentum are conserved. As in the case of the Compton effect,  when a photon with some energy collides with a stationary electron, some of the energy and momentum is transferred to the electron  but both energy and momentum are conserved in this elastic collision.so the correct option would be option B.

Cathode rays are high energetic electron beam.
Specific charge of cathode rays = e​/m_e​
Specific charge of positive rays =∣e∣/m​
Since the mass of positive rays is much larger than that of cathode rays, so the specific charge for positive rays is much smaller than that for cathode rays.

Photoelectrons will come out from the surface of the metal only if it gets enough energy during the irradiation the energy needed depends on the metal and the max energy that can be provided on the light.

Voltage across the electrodes of a cathode ray gun, V=500V
Charge of the electron=1.6x10^-19
Energy=eV
E= 1.6x10^-19 x 500
E=800 x 10^-19 J
E=8 x 10^-17 J

The kinetic energy (K.E.) of photoelectrons is calculated using the energy of the incident radiation, given by the formula:

• Incident Energy (E): E = hc/λ
• Where h = Planck's constant, c = speed of light, λ = wavelength
• For energy in electron volts (eV) and wavelength in Angstroms (Å), the formula simplifies to: E = 12400/λ Å

Given:

• Wavelength, λ = 1000 Å
• E = 12400 / 1000 Å
• Therefore, E = 12.4 eV

Since the work function is negligible, the K.E. of the photoelectrons is approximately the same as the incident energy: 12.4 eV.

Given, the maximum kinetic energy: K_max​=4eV
If V0​ be the stopping potential, then K_max​=eV₀​
⇒eV_0​=4eV 
⇒V₀​=4V

Photoelectric cell or photocell, device whose electrical characteristics (e.g., current, voltage, or resistance) vary when light is incident upon it. The most common type consists of two electrodes separated by a light-sensitive semiconductor material.
The photoelectric effect is the observation that many metals emit electrons when light shines upon them. Electrons emitted in this manner can be called photoelectrons.

A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

The maximum velocity of an electron emitted from the photocell is given as 3.75×10⁸ cm/s. To find the stopping potential, follow these steps:

• Use the formula for kinetic energy: KE = 0.5mv².
• Convert the velocity into metres per second: 3.75×10⁶ m/s.
• Calculate the kinetic energy of the electron: 
  KE = 0.5 × 9.11×10^-31 kg × (3.75×10⁶ m/s)².
• Convert the kinetic energy into electron volts (eV): 1 eV = 1.6×10^-19 J.
• The stopping potential, V, is given by the kinetic energy in eV, which is approximately 40 volts.

The maximum kinetic energy for the photoelectrons is 
E_max​=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
If V_o​ is the stopping potential then
eV₀​=h(c/λ)​−ϕ .....................(since, ν=c/λ​)
As per the problem, for incident light of 4000A^o, the stopping potential is 2V. When the wavelength of incident light is reduced to 3000A^o, then the stopping potential will increase to value more than 2V(as per the above equation).

The number of electrons emitted per second is observed to be directly proportional to the intensity of light. “Ok, so light is a wave and has energy. It takes electrons out of a metal, what is so special about that!” First of all, when the intensity of light is increased, we should see an increase in the photocurrent (number of photoelectrons emitted). Right?
As we see, this only happens above a specific value of frequency, known as the threshold frequency. Below this threshold frequency, the intensity of light has no effect on the photocurrent! In fact, there is no photocurrent at all, however high the intensity of light is.

The graph between the photoelectric current and the intensity of light is a straight line when the frequency of light used is above a specific minimum threshold value.

The photoelectric effect can only be explained by assuming that light consists of quanta or photons, as proposed by Einstein. According to the photoelectric effect, light behaves as a particle and transfers energy in discrete amounts (quanta or photons) to electrons in a material, causing them to be emitted from the surface. This is not explained by the wave nature of light, which is why options B, C, and D are incorrect.

The Stopping potential =2.5V.
or, Kinetic energy=2.5eV.
We know that,
Incident energy =work function + Kinetic energy.
To get incident energy in e.V,
We also know that,
The Incident energy =12400/λ Å
Incident energy=work function + kinetic energy.
12400/3000 = work function + 2.5e.v.
4.13-2.5 = work function
work function=1.64 e.V

When intensity of incident photons increases, the number of electrons emitted from the surface increases due to which current increases. Frequency of incident photons only limits the maximum kinetic energy of the photoelectron emitted and so the current is independent of frequency of photon.

Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons with the maximum kinetic energy K_max​ so that K_max​=eV_0​
For a given frequency of the incident radiation, the stopping potential is independent of its intensity.

Relation between kinetic energy and frequency is K.E+ϕ=hν.
Here h is planck's constant, ϕ=constant (Work function)
According to the above equation of photoelectric effect, as we increase frequency of photon, kinetic energy of photoelectron increases. Because the frequency of incident light is directly proportional to kinetic energy.
Hence option C is correct.
 

A photon is a particle of light which essentially is a packet of electromagnetic radiation. The energy of the photon depends on its frequency (how fast the electric field and magnetic field wiggle). The higher the frequency, the more energy the photon has. Of course, a beam of light has many photons. This means that really intense red light (lots of photons, with slightly lower energy) can carry more power to a given area than less intense blue light (fewer photons with higher energy).
 The speed of light (c) in a vacuum is constant. This means more energetic (high frequency) photons like X-rays and gamma rays travel at exactly the same speed as lower energy (low frequency) photons, like those in the infrared. As the frequency of a photon goes up, the wavelength goes down, and as the frequency goes down, the wavelength increases.
 

The number of electrons ejected can be measured as a function of intensity.
Intensity is equal to energy per unit time per unit area.
Keeping frequency constant, increasing intensity will increase energy thus increase of ejected electrons.
Thus,
Number of ejected photoelectrons will increase with increase in intensity of light. 

Given - φ = 3.3 × 10⁻¹⁹ J and h = 6.6 × 10⁻³⁴ J·s

We know that, Φ = hν₀ = hc / λ₀
Where ν₀ = threshold frequency, λ₀ = threshold wavelength, and c = speed of light

The work function is written as –
ν₀ = φ / h
⇒ ν₀ = (3.3 × 10⁻¹⁹) / (6.6 × 10⁻³⁴) = 0.5 × 10¹⁵ = 5 × 10¹⁴ Hz

The dual nature of light is demonstrated by the following phenomena:

• Diffraction: Light spreads out when it passes through a small opening or around an obstacle, demonstrating wave-like behaviour.
• Photoelectric Effect: Light ejects electrons from a material when it shines on it, showing particle-like properties.

These examples highlight how light behaves both as a wave and as a particle.