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JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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The increase in the energy of an electron for changing its de-Broglie wavelength from

1\,\mathop \mathrm{A}\limits^\mathrm{o} \mathrm{ }\,\,\mathrm{to}\,\,\mathrm{ }0.5\,\mathop \mathrm{A}\limits^\mathrm{o} \,

will be-

View answer and explanation

Correct option: B

0.45 keV

The correct answer is 0.45 keV because the change in energy is calculated from the change in de Broglie wavelength using the kinetic energy formula.

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The electrons are emitted in the photoelectric effect from a metal surface What is the de Broglie wavelength of a bullet of mass 0.04 kg travelling at the speed of 1 km/s? Photoelectrons are emitted from a metal surface when UV light of wavelength

\lambda=300 \mathrm{~nm}

is incident on it. The minimum negative potential required to stop the emission of electrons is

0.54 \mathrm{~V}

. Calculate:
(i) the energy of the incident photons
(ii) the maximum kinetic energy of the photoelectrons emitted. An electron is accelerated by a potential difference of V volt to acquire kinetic energy equal to maximum kinetic energy of ejected photoelectron from a metal of work function 3.2 eV, using a light of wavelength

2000\,\mathop \mathrm{A}\limits^ \circ \,

Value of V is Two monochromatic beams

A

B

of equal intensity

I

, hit a screen. The number of photons hitting the screen by beam

A

is twice that by beam

B

. Then, what can inference you about their frequencies? A 5 W source emits monochromatic light of wavelength

5000\mathop A\limits^ \circ

. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0m, the number of photoelectrons liberated will be reduced by a factor of