An electron of mass 'm' with an initial velocity v=v0i(v0>0) enters an electric field E=-E0k If the initial de Broglie…

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Q8 jee_main_2025_28_jan_evening Wave Particle Duality

Which of the following phenomena can not be explained by wave theory of light? [cite: 52-53]

A. Reflection of light
B. Diffraction of light
C. Refraction of light
D. Compton effect

Solution

Core Logic

Reflection, Refraction, and Diffraction can all be fully explained using Huygens' principle and wave theory paths [cite: 54, 47, 66].
Compton effect involves the scattering of an X-ray photon by an electron, demonstrating explicit momentum conversion. This process requires treating light strictly as localized particle packets (photons) and cannot be captured by continuous classical wave formulations.

Step 1: Conclusion

Hence, the Compton effect is the correct answer as it relies entirely on the particle nature of electromagnetic waves.

Pattern Recognition

Phenomena such as the Photoelectric Effect, Compton Scattering, and Blackbody Radiation serve as absolute foundational evidence for the particle/quantum layout of radiation, while Interference, Diffraction, and Polarization confirm wave configurations.

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Class 12 Physics: Dual Nature of Radiation and Matter

Q jee_main_2025_29_jan_morning Photoelectric Effect

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:

A. $(A) is false but (R) is true.$ $\text{(A) is false but (R) is true.}$
B. $(A) is true but (R) is false.$ $\text{(A) is true but (R) is false.}$
C. $Both (A) and (R) are true and (R) is the correct explanation of (A).$
D. $Both (A) and (R) are true but (R) is not the correct explanation of (A).$

Solution

Related Formula

eV₀ = hν - φ₀

Core Logic

Assertion (A) is true because applying a negative stopping potential decelerates the emitted photoelectrons and drops the output current down to zero. Reason (R) is true because the stopping potential $V₀ = ((h)/(e))ν - (φ₀)/(e)$ is linear with frequency $ν$ $\nu$. However, the linearity of $V₀$ $V_0$ vs frequency does not explain the physical mechanism behind why a negative potential stops electron emission.

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Class 12 Physics: Dual Nature of Radiation and Matter

Q jee_main_2025_29_jan_morning de Broglie Wavelength

If

λ

and K are de Broglie Wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be :-

A.
B.
C.
D.

Solution

Related Formula

$λ = h\sqrt(2mK)$ $\lambda = \frac{h}{\sqrt{2mK}}$

λ² = ((h²)/(2m)) (1)/(K)

Core Logic

Rearranging the de Broglie equation displays a parabolic relationship when evaluating squared attributes or corresponding axes coordinates. Given standard $λ$ $\lambda$ vs $1\sqrt(K)$ $\frac{1}{\sqrt{K}}$ layout tracking, it exhibits an upward facing parabolic behavior matching option (2) layout.

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Class 12 Physics: Dual Nature of Radiation and Matter

Q41 jee_main_2024_01_february_morning de Broglie Wavelength

The de Broglie wavelengths of a proton and an

α

particle are

λ

and

2λ

respectively. The ratio of the velocities of proton and

α

particle will be:

A. 1 : 8
B. 1 : 2
C. 4 : 1
D. 8 : 1

Solution

Related Formula

de Broglie wavelength relationship to velocity:

λ = (h)/(p) = (h)/(mv) v = (h)/(mλ)

Core Logic

Let mass of proton be $mₚ$ $m_p$ and mass of $α$ $\alpha$ particle be $m_α = 4mₚ$ $m_\alpha = 4m_p$. Given wavelengths: $λₚ = λ$ $\lambda_p = \lambda$, $λ_α = 2λ$ $\lambda_\alpha = 2\lambda$.

Set up ratios:

(vₚ)/(v_α) = (m_α)/(mₚ) \times (λ_α)/(λₚ)

Step 1: Substitute Ratios

$(vₚ)/(v_α) = 4 \times (2λ)/(λ) = 4 \times 2 = 8

Hence, the velocity ratio is$ Hence, the velocity ratio is $8:1. Pattern Recognition Remember the standard mass ratio:$. Pattern Recognition Remember the standard mass ratio: $m_\alpha \approx 4 m_p$. Inversely proportional components mean velocity amplifies significantly. Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

Q31 jee_main_2024_29_january_evening Photon Theory of Light

Two sources of light emit with a power of

200 W

. The ratio of number of photons of visible light emitted by each source having wavelengths

300 nm

and

500 nm

respectively, will be:

A. $1:5$ $1:5$
B. $1:3$ $1:3$
C. $5:3$ $5:3$
D. $3:5$ $3:5$

Solution

Related Formula

The power $P$ $P$ of a light source emitting $n$ $n$ photons per second of wavelength $λ$ $\lambda$ is given by:

P = n \cdot (hc)/(λ)

where:

$h$ $h$ is Planck's constant
$c$ $c$ is the speed of light

Core Logic

Since both light sources emit with the same power ( $P = 200 W$ $P = 200\text{ W}$), we can relate the number of photons emitted per second for each wavelength:

n₁ (hc)/(λ₁) = n₂ (hc)/(λ₂)

Cancelling out the constant terms $h$ $h$ and $c$ $c$, we get:

(n₁)/(λ₁) = (n₂)/(λ₂) (n₁)/(n₂) = (λ₁)/(λ₂)

Thus, the ratio of the number of photons emitted is directly proportional to their wavelengths.

Step 1: Substitute the Values

Given values:

$λ₁ = 300 nm$ $\lambda_1 = 300\text{ nm}$
$λ₂ = 500 nm$ $\lambda_2 = 500\text{ nm}$

Substituting these values into the ratio equation:

(n₁)/(n₂) = (300)/(500) = (3)/(5)

Thus, the ratio is $3:5$ $3:5$.

Pattern Recognition

Shortcut: For equal power outputs, the photon emission rate $n$ $n$ is directly proportional to the wavelength $λ$ $\lambda$. Therefore, ratio of photons $n₁ : n₂ = λ₁ : λ₂ = 300 : 500 = 3:5$ directly.

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Class 12 Physics: Dual Nature of Radiation and Matter

An electron of mass 'm' with an initial velocity v=v₀ i(v₀>0)$\vec{v}=v_{0}\hat{i}(v_{0}>0)$ enters an electric field E=-E₀ k$\vec{E}=-E_{0}\hat{k}$ If the initial de Broglie wavelength is λ₀$\lambda_{0}$, the value after time t would be :-

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculating Velocity Magnitude and Final Wavelength

Pattern Recognition

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More Dual Nature of Radiation and Matter Previous-Year Questions — Page 5

Solution

Core Logic

Step 1: Conclusion

Pattern Recognition

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Solution

Related Formula

Core Logic

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Solution

Related Formula

Core Logic

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Solution

Related Formula

Core Logic

Step 1: Substitute Ratios

Pattern Recognition

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Substitute the Values

Pattern Recognition

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An electron of mass 'm' with an initial velocity $v=v₀ i(v₀>0)$ $\vec{v}=v_{0}\hat{i}(v_{0}>0)$ enters an electric field $E=-E₀ k$ $\vec{E}=-E_{0}\hat{k}$ If the initial de Broglie wavelength is $λ₀$ $\lambda_{0}$, the value after time t would be :-