Q: The analysis of the Davisson-Germer results made use of:

Bragg's law of crystal diffraction. Applying 2d\sin\theta = n\lambda to the observed maximum gave a wavelength matching the de Broglie prediction.

Q: In the Davisson-Germer experiment, the accelerating voltage at which a pronounced maximum was observed was about:

54\text{ V}. At this voltage the de Broglie wavelength matched the Bragg condition for the nickel crystal.

Q: The de Broglie wavelength of an electron accelerated through 54\text{ V} is approximately:

1.67\text{ \AA}. \lambda = \dfrac{12.27}{\sqrt{54}} = \dfrac{12.27}{7.35} \approx 1.67\text{ \AA}, the value found by Davisson and Germer.

Q: The de Broglie wavelength of an electron accelerated through 150\text{ V} is approximately:

1.0\text{ \AA}. \lambda = \dfrac{12.27}{\sqrt{150}} = \dfrac{12.27}{12.25} \approx 1.0\text{ \AA}.

Question 1

The Davisson-Germer experiment demonstrated the:

Accepted Answer

Wave nature of electrons. Electrons scattered from a nickel crystal gave a diffraction maximum, confirming de Broglie's hypothesis.

Question 2

In the Davisson-Germer experiment, electrons were diffracted by:

Accepted Answer

A single crystal of nickel. The regular atomic spacing of the crystal acts as a natural grating for wavelengths of the order of an angstrom.

Question 3

The analysis of the Davisson-Germer results made use of:

Accepted Answer

Bragg's law of crystal diffraction. Applying 2d\sin	heta = n\lambda to the observed maximum gave a wavelength matching the de Broglie prediction.

Question 4

An electron microscope achieves much higher resolving power than an optical microscope because electrons have:

Accepted Answer

A very much shorter de Broglie wavelength than visible light. Resolution improves as the wavelength shortens, and accelerated electrons can reach well below 0.01	ext{ nm}.

Question 5

In an electron microscope, the electron beam is focused by:

Accepted Answer

Electric and magnetic fields acting as lenses. Charged particles are deflected by fields, so suitably shaped fields behave exactly like lenses.

Question 6

Electron diffraction was also demonstrated independently by:

Accepted Answer

G. P. Thomson. He passed electrons through thin metal foils and obtained ring patterns like those from X-rays.

Question 7

In the Davisson-Germer experiment, the accelerating voltage at which a pronounced maximum was observed was about:

Accepted Answer

54	ext{ V}. At this voltage the de Broglie wavelength matched the Bragg condition for the nickel crystal.

Question 8

The resolving power of a microscope improves when the wavelength used is:

Accepted Answer

Decreased. The limit of resolution is proportional to wavelength, which is the whole motivation for electron microscopy.

Question 9

The de Broglie wavelength of an electron accelerated through 54	ext{ V} is approximately:

Accepted Answer

1.67	ext{ \AA}. \lambda = \dfrac{12.27}{\sqrt{54}} = \dfrac{12.27}{7.35} \approx 1.67	ext{ \AA}, the value found by Davisson and Germer.

Question 10

The de Broglie wavelength of an electron accelerated through 150	ext{ V} is approximately:

Accepted Answer

1.0	ext{ \AA}. \lambda = \dfrac{12.27}{\sqrt{150}} = \dfrac{12.27}{12.25} \approx 1.0	ext{ \AA}.

Question 11

The de Broglie wavelength of an electron accelerated through 64	ext{ V} is approximately:

Accepted Answer

1.53	ext{ \AA}. \lambda = \dfrac{12.27}{8} \approx 1.53	ext{ \AA}.

Question 12

The de Broglie wavelength of an electron accelerated through 225	ext{ V} is approximately:

Accepted Answer

0.818	ext{ \AA}. \lambda = \dfrac{12.27}{15} \approx 0.818	ext{ \AA}.

Question 13

An electron beam is to have a de Broglie wavelength of 1	ext{ \AA}. The required accelerating voltage is approximately:

Accepted Answer

150	ext{ V}. From \lambda = \dfrac{12.27}{\sqrt{V}}, V = (12.27)^2 \approx 150	ext{ V}.

Question 14

The de Broglie wavelength of an electron accelerated through 50	ext{ V} is approximately:

Accepted Answer

1.74	ext{ \AA}. \lambda = \dfrac{12.27}{\sqrt{50}} = \dfrac{12.27}{7.07} \approx 1.74	ext{ \AA}.

Question 15

In a crystal of spacing 0.91	ext{ \AA}, a first-order Bragg maximum occurs at 65^\circ. The wavelength is approximately:

Accepted Answer

1.65	ext{ \AA}. \lambda = 2d\sin	heta = 2 	imes 0.91 	imes \sin 65^\circ \approx 1.65	ext{ \AA}.

Question 16

If the accelerating voltage in an electron microscope is quadrupled, the de Broglie wavelength:

Accepted Answer

Halves. \lambda \propto \dfrac{1}{\sqrt{V}}, so the resolving power improves by a factor of two.

Question 17

An electron microscope uses electrons of wavelength about 0.01	ext{ nm} while an optical microscope uses light of 500	ext{ nm}. The improvement in resolving power is roughly:

Accepted Answer

5 	imes 10^4 times. Resolving power \propto \dfrac{1}{\lambda}, so the ratio is \dfrac{500}{0.01} = 5 	imes 10^4.

Question 18

The de Broglie wavelength of an electron accelerated through 900	ext{ V} is approximately:

Accepted Answer

0.41	ext{ \AA}. \lambda = \dfrac{12.27}{30} \approx 0.41	ext{ \AA}.

Question 19

Two electron beams are accelerated through 100	ext{ V} and 400	ext{ V}. The ratio of their de Broglie wavelengths is:

Accepted Answer

2 : 1. \lambda \propto \dfrac{1}{\sqrt{V}}, so the ratio is \sqrt{400} : \sqrt{100} = 2 : 1.

Question 20

The electron microscope is preferred for viewing viruses because it offers:

Accepted Answer

A much smaller limit of resolution than an optical microscope. Objects far smaller than the wavelength of visible light simply cannot be resolved optically.

Davisson-Germer & Electron Microscope — MH-CET Physics MCQs with Solutions

Questions with solutions