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JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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Analyzing +4 | -1

Assertion: de-Broglie wavelength of an electron accelerated through a potential difference of

{V}

volt is

\lambda=\frac{12.72}{\sqrt{V}}{A}

Reason: de-Broglie wavelength of an electron is given by

\lambda=\frac{h}{m v}

.

View answer and explanation

Correct option: D

Assertion is incorrect, reason is correct.

The assertion is incorrect, and the reason is correct. The correct formula for de-Broglie wavelength is

\lambda=\frac{{12.27}}{{\sqrt{{V}}}} {A}

, not

\lambda=\frac{{12.72}}{{\sqrt{{V}}}} {A}

.

More Questions from DUAL NATURE OF RADIATION AND MATTER

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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