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bekas [8.4K]
3 years ago
6

A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electr

ons and the protons have the same speed. Which one of the following correctly describes the beam that experiences the greatest amount of diffraction?
(a) The electrons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
(c) The protons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(d) The protons, because they have the larger momentum and, hence, the smaller de Broglie wavelength
(e) Both beams experience the same amount of diffraction, because the electrons and protons have the same de Broglie wavelength.
Physics
1 answer:
ryzh [129]3 years ago
7 0

Answer:

(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength

Explanation:

de Broglie wavelength λ = h / m v

Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .

for electron , momentum is less so  . Therefore de Broglie wavelength λ will be more for electrons .

Amount of diffraction that is angle of diffraction is proportional to λ

Therefore electrons having greater de Broglie wavelength will show greater diffraction.

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A baseball with a mass of 0.80 kg is given an acceleration of 20.00 m/s. How much net force was applied to the ball
dybincka [34]

a x m = f

.80 x 20 = 16

4 0
3 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place
Semmy [17]

Answer:

Ans= 9

See attached picture for clearer solution.

Explanation:

The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .  

The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .  

The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .  

The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .  

The ratio of the largest to the smallest net force = 9/4*2/2  /  1/4 2/2 . = 9

5 0
3 years ago
Which tool would you use to measure the amount of rainfall?
zheka24 [161]

Answer:

a

Explanation:

5 0
3 years ago
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