Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. 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 threshold frequency. In the light of the above statements, choose the correct answer from the options given below

Einstein's photoelectric equation: where: * Vs = stopping potential * u = frequency of light * phi = work function * Intensity formula: I = (n h u/A cdot t) (where n is rate of photons). * **Assertion Analysis:** Stopping potential Vs depends strictly linearly on frequency u and work function phi. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given u > u0. Thus, Reason R is **true**. Stopping Potential rightarrow Frequency/Energy characteristic. Photo-current / Emission Rate rightarrow Photon Intensity/Flux counts. Class 12 Physics: Dual Nature of Radiation and Matter

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping p

Einstein's photoelectric equation: where: * Vs = stopping potential * u = frequency of light * phi = work function * Intensity formula: I = (n h u/A cdot t) (where n is rate of photons). * **Assertion Analysis:** Stopping potential Vs depends strictly linearly on frequency u and work function phi. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up th

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. 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 threshold frequency. In the light of the above statements, choose the correct answer from the options given below

Einstein's photoelectric equation: where: * Vs = stopping potential * u = frequency of light * phi = work function * Intensity formula: I = (n h u/A cdot t) (where n is rate of photons). * **Assertion Analysis:** Stopping potential Vs depends strictly linearly on frequency u and work function phi. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given u > u0. Thus, Reason R is **true**. Stopping Potential rightarrow Frequency/Energy characteristic. Photo-current / Emission Rate rightarrow Photon Intensity/Flux counts. Class 12 Physics: Dual Nature of Radiation and Matter

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping p

Einstein's photoelectric equation: where: * Vs = stopping potential * u = frequency of light * phi = work function * Intensity formula: I = (n h u/A cdot t) (where n is rate of photons). * **Assertion Analysis:** Stopping potential Vs depends strictly linearly on frequency u and work function phi. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up th

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In…

Year	2026	2025	2024	Total
Questions	7	16	8	31

Q47 jee_main_2026_28_january_morning de Broglie Wavelength

The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is

n : 1

. The value of n is ____. (Assume mass of proton = mass of neutron)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

λ = (h)/(p) = h\sqrt(2m \cdot KE)

Core Logic

Set up the ratio of wavelengths for the deuteron and the alpha particle based on their given kinetic energies and masses. Let mass of proton/neutron be $m$ $m$. Deuteron mass is $2m$ $2m$, Alpha particle mass is $4m$ $4m$.

Step 1: Ratio setup

(λd)/(λ_α) = \sqrt((m_α \cdot KE_α)/(md \cdot KEd))

Step 2: Value Substitution

Substitute $m_α = 4m$ $m_\alpha = 4m$, $md = 2m$ $m_d = 2m$, $KE_α = 2E$ $KE_\alpha = 2E$, and $KEd = E$ $KE_d = E$:

= \sqrt((4m \cdot 2E)/(2m \cdot E))

= \sqrt((8mE)/(2mE)) = \sqrt(4) = 2

So, the ratio is $2 : 1$ $2 : 1$. Therefore, $n = 2$ $n = 2$.

Pattern Recognition

Memorize mass ratios for common particles: proton ( $m$ $m$), deuteron ( $2m$ $2m$), alpha ( $4m$ $4m$). Substitute directly into inverse-sqrt formula for $λ$ $\lambda$.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q36 jee_main_2026_28_january_evening Photon Energy

Number of photons of equal energy emitted per second by a

6 mW

laser source operating at

663 nm

is ____. (Given:

h = 6.63 \times 10⁻³⁴ J.s

and

c = 3 \times 10⁸ m/s

)

A. $5 \times 10¹⁶$ $5 \times 10^{16}$
B. $5 \times 10¹⁵$ $5 \times 10^{15}$
C. $10 \times 10¹⁵$ $10 \times 10^{15}$
D. $2 \times 10¹⁶$ $2 \times 10^{16}$

Solution

Related Formula

P = n (hc)/(λ)

where: $P$ $P$ = Power of the laser $n$ $n$ = number of photons emitted per second $h$ $h$ = Planck's constant $c$ $c$ = speed of light $λ$ $\lambda$ = wavelength

Step 1: Extract Given Variables

$P = 6 mW = 6 \times 10⁻³ J/s$ $λ = 663 nm = 663 \times 10⁻⁹ m$ $h = 6.63 \times 10⁻³⁴ J.s$ $h = 6.63 \times 10^{-34} \text{ J.s}$ $c = 3 \times 10⁸ m/s$ $c = 3 \times 10^8 \text{ m/s}$

Step 2: Substitution and Calculation

6 \times 10⁻³ = n \times 6.63 \times 10⁻³⁴ \times 3 \times 10⁸663 \times 10⁻⁹

Notice that $6.63 / 663 = 10⁻²$ $6.63 / 663 = 10^{-2}$.

6 \times 10⁻³ = n \times 3 \times 10⁻²⁶ \times 10⁻²10⁻⁹

6 \times 10⁻³ = n \times 3 \times 10⁻¹⁹

n = 6 \times 10⁻³3 \times 10⁻¹⁹ = 2 \times 10¹⁶

Pattern Recognition

Always look for complementary numerical values like $663$ $663$ and $6.63$ $6.63$. It guarantees a clean power of 10 extraction.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q5 jee_main_2025_02_april_evening de-Broglie Wavelength

An electron with mass

m

with an initial velocity

(t = 0)

v = v₀ i

(

v₀ > 0

) enters a magnetic field

B = B₀ j

. If the initial de-Broglie wavelength at

t = 0

is

λ₀

then its value after time

t

would be:

A. $λ₀\sqrt(1 - (e² B₀² t²)/(m²))$
B. $λ₀\sqrt(1 + (e² B₀² t²)/(m²))$
C. $λ₀ \sqrt(1 + (e² B₀² t²)/(m²))$
D. $λ₀$ $\lambda_0$

Solution

Related Formula

Magnetic Force on a moving charge:

F = q( v \times B)

de-Broglie Wavelength:

λ = (h)/(p) = (h)/(m v)

where $p$ $p$ is the magnitude of momentum and $v$ $v$ is the speed.

Core Logic

Since the magnetic force $F$ $\vec{F}$ is always perpendicular to the velocity $v$ $\vec{v}$ of the electron at any instant:

W = \int F \cdot d r = 0

By the work-energy theorem, since work done by the magnetic field is zero, the kinetic energy (and thus the speed $v$ $v$) of the electron remains constant throughout its motion.

Since speed $v = v₀$ $v = v_0$ (constant), the magnitude of momentum $p = m v$ $p = m v$ remains constant over time.

Therefore, the de-Broglie wavelength remains unchanged:

λ(t) = λ₀

Pattern Recognition

Sees: Charge entering purely magnetic field. Trap: Resolving helical trajectories or cross products mathematically. Do not waste time computing components! Shortcut: A magnetic field can ONLY change the direction of velocity, NEVER the magnitude (speed). Since de-Broglie wavelength depends solely on the magnitude of momentum ( $p = mv$ $p = mv$), it must remain constant.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter Class 12 Physics: Moving Charges and Magnetism

Q20 jee_main_2025_02_april_morning Photoelectric Effect

A monochromatic light is incident on a metallic plate having work function

φ

. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is: (Given: The magnitude of charge of an electron is

e

and mass is

m

,

h

is Planck's constant and

c

is velocity of light. Take the magnetic field exists throughout the path of electron)

A. $\sqrt(2m((hc)/(λ) - φ)) / eB$
B. $\sqrt(m((hc)/(λ) - φ)) / eB$
C. $\sqrt(8m((hc)/(λ) - φ)) / eB$
D. $2 \sqrt(m((hc)/(λ) - φ)) / eB$

Solution

Related Formula

K = (hc)/(λ) - φ

p = 2m K

R = (p)/(eB)

$d = 2R$ $d = 2R$

Core Logic

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is:

K = (hc)/(λ) - φ

The momentum $p$ $p$ corresponding to this kinetic energy is:

p = 2m K = \sqrt(2m ((hc)/(λ) - φ))

The electron is emitted normally to the plate and enters a perpendicular constant magnetic field $B$ $B$. It describes a circular arc (semicircle) and hits back the plate at point B. The distance between A and B is the diameter of this circular trajectory:

dAB = 2R = 2 ( (p)/(eB) ) = 2\sqrt(2m((hc)/(λ) - φ))eB

To align this with the options, move the factor of 2 inside the square root ( $2 = \sqrt(4)$ $2 = \sqrt{4}$):

dAB = \sqrt(4 \times 2m((hc)/(λ) - φ))eB = \sqrt(8m((hc)/(λ) - φ))eB

Step 1: Final Conclusion The distance between points A and B is$ Step 1: Final Conclusion The distance between points A and B is $\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)} / eB. Pattern Recognition When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter$. Pattern Recognition When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter $2R = 2\frac{p}{qB}. Taking coefficients inside square roots converts$. Taking coefficients inside square roots converts $2 \sqrt{2x}to$ to $\sqrt{8x}$. Chapter Mix Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Moving Charges and Magnetism

Q21 jee_main_2025_08_april_evening de-Broglie Wavelength

An electron is released from rest near an infinite non-conducting sheet of uniform charge density

-σ

. The rate of change of de-Broglie wavelength associated with the electron varies inversely as

nth

power of time. The numerical value of

n

is

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

λ = (h)/(p)

p = m v = m (at)

a = (e E)/(m) = e σ2mε₀

where, $λ$ $\lambda$ = de-Broglie wavelength $p$ $p$ = linear momentum $a$ $a$ = acceleration of the electron in the uniform electric field $E$ $E$ $t$ $t$ = time elapsed since release

Core Logic

Since the electron starts from rest ( $u = 0$ $u = 0$), its velocity $v$ $v$ at any time $t$ $t$ is:

$v = at$ $v = at$

Thus, the momentum is $p = m v = m a t$ $p = m v = m a t$.

Substitute this into the de-Broglie wavelength equation:

λ(t) = (h)/(m a t)

Now, compute the rate of change of wavelength with respect to time:

(dλ)/(dt) = (d)/(dt) ( (h)/(ma) t⁻¹ ) = -(h)/(ma) t⁻²

This shows that:

| (dλ)/(dt) | \propto (1)/(t²)

Comparing this with the given statement (varies inversely as $nth$ $n^{\text{th}}$ power of time):

$n = 2$ $n = 2$

Pattern Recognition

Sees: "Uniform electric field" + "de-Broglie wavelength rate of change" → Wavelength $λ \propto t⁻¹$ $\lambda \propto t^{-1}$. Shortcut: Since $λ \propto (1)/(t)$ $\lambda \propto \frac{1}{t}$, its derivative must scale as $(dλ)/(dt) \propto (1)/(t²)$ . Thus, $n = 2$ $n = 2$ directly from basic power-rule differentiation! ✓

Chapter Mix

Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Electrostatics

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Dual Nature of Radiation and Matter Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Ratio setup

Step 2: Value Substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Extract Given Variables

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Step 1: Final Conclusion

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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