JEE Main 2024 (Online) 27th January Evening Shift | Dual Nature of Radiation Question 23 | Physics

The de Broglie wavelength of an electron having kinetic energy $\mathrm{E}$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its de-Broglie wavelength will be :

$\sqrt{2} \lambda$

$2 \lambda$

$\frac{\lambda}{2}$

$\frac{\lambda}{\sqrt{2}}$

JEE Main 2023 (Online) 13th April Evening Shift

MCQ (Single Correct Answer)

-1

Given below are two statements:

Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.

Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.

In the light of above statements, choose the correct answer form the options given below

Both Statement I and Statement II are true

Statement I is true but statement II is false

Statement I is false but statement II is true

Both Statement I and Statement II are false

JEE Main 2023 (Online) 13th April Morning Shift

MCQ (Single Correct Answer)

-1

The difference between threshold wavelengths for two metal surfaces $$\mathrm{A}$$ and $$\mathrm{B}$$ having work function $$\phi_{A}=9 ~\mathrm{eV}$$ and $$\phi_{B}=4 \cdot 5 ~\mathrm{eV}$$ in $$\mathrm{nm}$$ is:

$$\{$$ Given, hc $$=1242 ~\mathrm{eV} \mathrm{nm}\}$$

264

138

540

276

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