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Romashka [77]
3 years ago
9

A proton, an electron, and a helium nucleus all move at speed v. Rank their de Broglie wavelengths from largest to smallest. Wri

te any relevant formulas you use below. a) helium nucleus, proton, electron
b) helium nucleus, electron, proton
c) electron, proton, helium nucleus
d) proton, electron, helium nucleus
e) proton, helium nucleus, electron
Physics
1 answer:
MatroZZZ [7]3 years ago
6 0

Answer:

The correct option is 'c':electron,proton,helium nucleus

Explanation:

The De-Broglie's wavelength of particle is given by

\lambda =\frac{h}{mv}

Thus we can see that wavelength is inversely related to mass of the particle since 'h' (Plank's constant) and velocity is same for all the particles  

Thus we conclude that the the lightest particle will have the most wavelength

Electron being the lightest of the 3 particles will have the largest wavelength thus the correct option is 'c'. Since electron has the largest wavelength followed by proton and the least wavelength among the 3 is of helium.

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A mass m is attached to a spring which is held stretched a distance x by a force f (fig. 7-28), and then released. the spring co
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Spring is stretched by force f to distance "x"

now here by force balance we can say

f = kx

k = \frac{f}{x}

now here we will we say that energy stored in the spring will convert into kinetic energy

\frac{1}{2} kx^2 = \frac{1}{2}mv^2

\frac{f}{x} (x^2} = mv^2

now solving above equation we will have

v =\sqrt{ \frac{fx}{m}}

PART 2)

now for half of the extension again we can use energy conservation

\frac{1}{2}kx^2 - \frac{1}{2}k(x/2)^2 = \frac{1}{2} mv^2

\frac{3}{4}kx^2 = mv^2

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v = \sqrt{\frac{3fx}{4m}}

5 0
3 years ago
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7 0
3 years ago
Two billiard balls having the same mass and speed roll directly toward each other. What is their combined momentum after they me
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Their combined momentum after they meet is 0 .
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Photosynthesis and respiration are best described as
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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find
sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge q_{1}=9\times10^{-6}\ C at origin

Second charge q_{2}=6\times10^{-6}\ C

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

E=\dfrac{kq}{r^2}

0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}

\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}

\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1

\dfrac{1}{d}=1.82

d=\dfrac{1}{1.82}

d=0.55\ m

Hence, The coordinates of the point is (0,0.55).

3 0
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