The work function of a substance is 3.0 ~eV. The longest wavelength of light that can cause the emission of…

Year	2026	2025	2024	Total
Questions	7	16	8	31

Q37 jee_main_2026_21_jan_morning Photoelectric Effect

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

The work function of a substance is 3.0 ~eV$3.0 \mathrm{~eV}$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Wavelength

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_jan_morning

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

The work function of a substance is $3.0 ~eV$ $3.0 \mathrm{~eV}$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately: