COMEDK 2025 Afternoon Shift | Dual Nature of Radiation Question 3 | Physics

de Broglie wave length associated with an electron accelerated through a potential difference ' V ' is ' $\lambda$ '. If the accelerating potential is halved, what will be the new wave length associated with the charged particle?

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

$\frac{\lambda}{2}$

$2 \lambda$

$\sqrt{2} \boldsymbol{\lambda}$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

-0

Which of the following statement is wrong regarding the photo electric effect.

Photo electric current is directly proportional to the intensity of the incident radiation

The maximum kinetic energy of the electrons emitted depends on the intensity of the incident radiation

The maximum kinetic energy of the electrons emitted depends on the frequency of the incident radiation

The value of the stopping potential increases with the increase in frequency of the incident radiation.

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

A particle of mass $M$ at rest decays into masses $m_1$ and $m_2$ with non-zero velocities. The ratio of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ of the particles is

$\frac{m_2}{m_1}$

$\frac{\sqrt{m_1}}{\sqrt{m_2}}$

$\frac{m_1}{m_2}$

$1: 1$

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

Emission of electrons from a metal plate illuminated with monochromatic electromagnetic radiation will always take place provided

The plate is positively charged

The plate is negatively charged

The radiation is sufficiently intense

The work function of the plate is less than the energy of a single photon and the plate is uncharged

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