For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (nu) of the incident photons as shown in figure. The slope of the graph gives

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by Ek = hnu - phi, where h is Planck's constant, nu is the incident frequency, and phi is the work function. This represents a straight line equation of the form y = mx + c, where y = Ek and x = nu. Comparing Ek = hnu - phi with y = mx + c, we get the slope: Thus, the slope of the graph gives Planck's constant. If the y-axis is K.E., the slope is h. If the y-axis is Stopping Potential V0, the equation is eV0 = hnu - phi V0 = (h/e)nu - (phi/e), and the slope would be h/e. Class 12 Physics: Dual Nature of Radiation and Matter

For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (nu) of the incident photons as shown in figure. The slope of the graph gives

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by Ek = hnu - phi, where h is Planck's constant, nu is the incident frequency, and phi is the work function. This represents a straight line equation of the form y = mx + c, where y = Ek and x = nu. Comparing Ek = hnu - phi with y = mx + c, we get the slope: Thus, the slope of the graph gives Planck's constant. If the y-axis is K.E., the slope is h. If the y-axis is Stopping Potential V0, the eq

For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (nu) of the incident photons as shown in figure. The slope of the graph gives

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by Ek = hnu - phi, where h is Planck's constant, nu is the incident frequency, and phi is the work function. This represents a straight line equation of the form y = mx + c, where y = Ek and x = nu. Comparing Ek = hnu - phi with y = mx + c, we get the slope: Thus, the slope of the graph gives Planck's constant. If the y-axis is K.E., the slope is h. If the y-axis is Stopping Potential V0, the equation is eV0 = hnu - phi V0 = (h/e)nu - (phi/e), and the slope would be h/e. Class 12 Physics: Dual Nature of Radiation and Matter

For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (nu) of the incident photons as shown in figure. The slope of the graph gives

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by Ek = hnu - phi, where h is Planck's constant, nu is the incident frequency, and phi is the work function. This represents a straight line equation of the form y = mx + c, where y = Ek and x = nu. Comparing Ek = hnu - phi with y = mx + c, we get the slope: Thus, the slope of the graph gives Planck's constant. If the y-axis is K.E., the slope is h. If the y-axis is Stopping Potential V0, the eq

For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency…

Year	2026	2025	2024	Total
Questions	7	16	8	31

Q37 jee_main_2026_21_jan_morning Photoelectric Effect

A light wave described by

E = 60[ (3 \times 10¹⁵t) + (12 \times 10¹⁵t)]

(in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. (

h = 6.6 \times 10⁻³⁴ J\dots

. and

e = 1.6 \times 10⁻¹⁹C

)

A. 5.1
B. 3.8
C. 6.0
D. 7.8

Solution

Related Formula

Kmax = hνmax - φ₀

v = (ω)/(2π)

Core Logic

The light wave consists of two frequencies governed by $ω₁$ $\omega_1$ and $ω₂$ $\omega_2$. $ω₁ = 3 \times 10¹⁵ rad/s$ $ω₂ = 12 \times 10¹⁵ rad/s$

The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to $ω₂ = 12 \times 10¹⁵ rad/s$ .

Step 1: Calculate Photon Energy

Frequency $νmax = (ω₂)/(2π) = 12 \times 10¹⁵2 \times 3.14 \approx 1.91 \times 10¹⁵ Hz$ .

Energy of this photon:

Ephoton = hν = (6.6 \times 10⁻³⁴) \times (1.91 \times 10¹⁵) = 1.26 \times 10⁻¹⁸ J

Convert this energy to eV:

Emax = 1.26 \times 10⁻¹⁸1.6 \times 10⁻¹⁹ \approx 7.87 eV \approx 7.9 eV

Step 2: Calculate Maximum Kinetic Energy

Using Einstein's photoelectric equation:

Kmax = Emax - φ₀

Kmax = 7.9 - 2.8 = 5.1 eV

Pattern Recognition

When a wave has multiple frequency components ( $E = E₁ ω₁ t + E₂ ω₂ t$ ), the $Kmax$ $K_{\text{max}}$ is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q48 jee_main_2026_21_jan_evening De Broglie Wavelength

A particle having electric charge

3 \times 10⁻¹⁹ C

and mass

6 \times 10⁻²⁷ kg

is accelerated by applying an electric potential of

1.21 V

. Wavelength of the matter wave associated with the particle is

α \times 10⁻¹² m

. The value of

α

is ________. (Take Planck's constant

= 6.6 \times 10⁻³⁴ J\dots

)

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

λ = (h)/(p) = h\sqrt(2mK)

$K = qV$ $K = qV$

Core Logic

For a charged particle accelerated through a potential difference $V$ $V$, the de Broglie wavelength is:

λ = h\sqrt(2mqV)

Step 1: Value Substitution

λ = 6.6 \times 10⁻³⁴ 2 \times 6 \times 10⁻²⁷ \times 3 \times 10⁻¹⁹ \times 1.21

λ = 6.6 \times 10⁻³⁴ 36 \times 10⁻⁴⁶ \times 1.21

Step 2: Arithmetic Evaluation

Evaluate the square root:

\sqrt(36) \times \sqrt(1.21) \times 10⁻⁴⁶ = 6 \times 1.1 \times 10⁻²³ = 6.6 \times 10⁻²³

λ = 6.6 \times 10⁻³⁴6.6 \times 10⁻²³

λ = 10⁻¹¹ m

Step 3: Final Conclusion

We need to express this in the form $α \times 10⁻¹² m$ $\alpha \times 10^{-12} \text{ m}$:

10⁻¹¹ m = 10 \times 10⁻¹² m

Therefore, $α = 10$ $\alpha = 10$.

Pattern Recognition

The expression inside the radical always evaluates cleanly in JEE. $2 \times 6 \times 3 = 36$ $2 \times 6 \times 3 = 36$ and $1.21 = (1.1)²$ $1.21 = (1.1)^2$ are engineered to perfectly cancel the $6.6$ $6.6$ in the numerator.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q44 jee_main_2026_22_january_evening Photoelectric Effect and Threshold Frequency

Light is incident on a metallic plate having work function

110 \times 10⁻²⁰

J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ____ rad/s. (

h = 6.63 \times 10⁻³⁴

J.s)

A. $1.04 \times 10¹⁶$ $1.04 \times 10^{16}$
B. $1.04 \times 10¹³$ $1.04 \times 10^{13}$
C. $1.66 \times 10¹⁶$ $1.66 \times 10^{16}$
D. $1.66 \times 10¹⁵$ $1.66 \times 10^{15}$

Solution

Related Formula

$φ = h v$ $\phi = h v$

ω = 2π v = (2π φ)/(h)

Core Logic

Since kinetic energy of photoelectrons is zero ( $Kmax = 0$ $K_{\text{max}} = 0$), incident photon energy equals work function $φ$ $\phi$:

h v = φ v = (φ)/(h)

Calculating angular frequency $ω$ $\omega$:

ω = 2π v = (2π φ)/(h) = 2 \times 3.14 \times 110 \times 10⁻²⁰6.63 \times 10⁻³⁴

ω = 690.8 \times 10⁻²⁰6.63 \times 10⁻³⁴ \approx 1.04 \times 10¹⁶ ~rad/s

Step 1: Final Conclusion

The angular frequency of incident light is $1.04 \times 10¹⁶ ~rad/s$ $1.04 \times 10^{16} \mathrm{~rad/s}$.

Pattern Recognition

Zero kinetic energy $hν = φ$ $\implies h\nu = \phi$. Convert linear frequency $ν$ $\nu$ to angular frequency $ω = 2π φ / h$ $\omega = 2\pi \phi / h$.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q30 jee_main_2026_23_january_morning De Broglie Wavelength

The de Broglie wavelength of an oxygen molecule at

27°C

is

x \times 10⁻¹²

m. The value of x is (take Planck's constant =

6.63 \times 10⁻³⁴ J.s

, Boltzmann constant =

1.38 \times 10⁻²³ J/K

, mass of oxygen. Molecule =

5.31 \times 10⁻²⁶ kg

).

A. $26$ $26$
B. $24$ $24$
C. $30$ $30$
D. $20$ $20$

Solution

Related Formula

λ = h\sqrt(2mK)

K = (3)/(2)kT

λ = h\sqrt(3mkT)

Step 1: Substitute Values

Given values: $h = 6.63 \times 10⁻³⁴ J\dots$ $m = 5.31 \times 10⁻²⁶ kg$ $m = 5.31 \times 10^{-26} \text{ kg}$ $k = 1.38 \times 10⁻²³ J/K$ $k = 1.38 \times 10^{-23} \text{ J/K}$ $T = 27°C = 27 + 273 = 300 K$

Substitute these into the wavelength equation:

λ = 6.63 \times 10⁻³⁴ 3 \times 5.31 \times 10⁻²⁶ \times 1.38 \times 10⁻²³ \times 300

Step 2: Simplify Calculation

λ = 6.63 \times 10⁻³⁴ 3 \times 300 \times 5.31 \times 1.38 \times 10⁻⁴⁹

λ = 6.63 \times 10⁻³⁴ 900 \times 7.3278 \times 10⁻⁴⁹

λ = 6.63 \times 10⁻³⁴ 6595.02 \times 10⁻⁴⁹

λ = 6.63 \times 10⁻³⁴ 659.502 \times 10⁻⁴⁸

λ = 6.63 \times 10⁻³⁴25.68 \times 10⁻²⁴ \approx 0.258 \times 10⁻¹⁰ m

Step 3: Match with Format

λ = 25.8 \times 10⁻¹² m

Rounding to nearest integer, $x = 26$ $x = 26$.

Pattern Recognition

Sees: "de Broglie wavelength" + "gas molecule" + "temperature" → Immediately use $λ = h/\sqrt(3mkT)$ $\lambda = h/\sqrt{3mkT}$. Watch for temperature conversion to Kelvin.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter Class 11 Physics: Kinetic Theory

Q33 jee_main_2026_24_january_evening Photoelectric Effect

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ____ m.

(h = 6.63\times 10⁻³⁴J.s,e = 1.6\times 10⁻¹⁹C,c = 3\times 10⁸m / s)

A. $2.9 \times 10⁻⁸$ $2.9 \times 10^{-8}$
B. $2.2 \times 10⁻⁸$ $2.2 \times 10^{-8}$
C. $3.1 \times 10⁻⁷$ $3.1 \times 10^{-7}$
D. $2.5 \times 10⁻⁷$ $2.5 \times 10^{-7}$

Solution

Related Formula

eVₛ = (hc)/(λ) - φ

Core Logic

For the first light:

e(3.2) = (hc)/(λ) - φ (1)

For the second light (wavelength $2λ$ $2\lambda$):

e(0.7) = (hc)/(2λ) - φ (2)

Photoelectric Effect diagram for Q33 - JEE Main 2026 Evening

Step 1: Solving the Equation

Subtract Equation (2) from Equation (1):

e(3.2) - e(0.7) = ( (hc)/(λ) - φ ) - ( (hc)/(2λ) - φ )

e(2.5) = (hc)/(2λ)

Step 2: Calculate Wavelength

2.5 = ((hc)/(e)) ((1)/(2λ))

Using $(hc)/(e) \approx 12400 eV\cdotAA$ :

2.5 = (12400)/(2λ)

λ = (12400)/(5) AA

λ = 2480 AA = 2.48 \times 10⁻⁷ m

Pattern Recognition

When you have two states of photoelectric effect for the same metal, subtracting the two stopping potential equations instantly eliminates the work function $φ$ $\phi$. Convert $hc/e$ $hc/e$ directly to $12400 eV\cdotAA$ $12400 \text{ eV}\cdot\text{\AA}$ to expedite calculation.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Solution & Explanation

Related Formula

Core Logic

Step 1: Identify the Slope

Pattern Recognition

Chapter Mix

Reference Study Guides

More Dual Nature of Radiation and Matter Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate Photon Energy

Step 2: Calculate Maximum Kinetic Energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Value Substitution

Step 2: Arithmetic Evaluation

Step 3: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Substitute Values

Step 2: Simplify Calculation

Step 3: Match with Format

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solving the Equation

Step 2: Calculate Wavelength

Pattern Recognition

Chapter Mix

More Dual Nature of Radiation and Matter Questions — jee_main_2024_30_january_evening

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