A photo-emissive substance is illuminated with a radiation of wavelength lambdai so that it releases electrons with…

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

More Dual Nature of Radiation and Matter Questions — jee_main_2025_07_april_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back